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哈希表专题</title><meta name="generator" content="Hexo 7.0.0"></head><body itemscope="" itemtype="http://schema.org/WebPage"><div id="loading"><div class="cat"><div class="body"></div><div class="head"><div class="face"></div></div><div class="foot"><div class="tummy-end"></div><div class="bottom"></div><div class="legs left"></div><div class="legs right"></div></div><div class="paw"><div class="hands left"></div><div class="hands right"></div></div></div></div><div id="container"><header id="header" itemscope="" itemtype="http://schema.org/WPHeader"><div class="inner"><div id="brand"><div class="pjax"><h1 itemprop="name headline">LeetCode - 哈希表专题</h1><div class="meta"><span class="item" title="创建时间：2022-05-03 21:06:21"><span class="icon"><i class="ic i-calendar"></i></span><span class="text">发表于</span><time itemprop="dateCreated datePublished" datetime="2022-05-03T21:06:21+08:00">2022-05-03</time></span><span class="item" title="本文字数"><span class="icon"><i class="ic i-pen"></i></span><span class="text">本文字数</span><span>31k</span><span class="text">字</span></span><span class="item" title="阅读时长"><span class="icon"><i class="ic i-clock"></i></span><span class="text">阅读时长</span><span>29 分钟</span></span></div></div></div><nav id="nav"><div class="inner"><div class="toggle"><div class="lines" aria-label="切换导航栏"><span class="line"></span><span class="line"></span><span class="line"></span></div></div><ul class="menu"><li class="item title"><a href="/" rel="start">Jiankychen</a></li></ul><ul class="right" id="rightNav"><li class="item theme"><i class="ic i-sun"></i></li><li class="item search"><i class="ic i-search"></i></li></ul></div></nav></div><div class="pjax" id="imgs"><ul><li class="item" style="background-image: url("https://i.imgtg.com/2023/03/09/YSj7p.jpg");"></li><li class="item" style="background-image: url("https://img.timelessq.com/images/2022/07/26/8fe50780c15461b629c9aeab5a7f2acd.jpg");"></li><li class="item" style="background-image: url("https://img.timelessq.com/images/2022/07/26/99fb5ff897a82984470abf5e2a235d94.jpg");"></li><li class="item" style="background-image: url("https://img.timelessq.com/images/2022/07/26/42bab566f107b9a16542343e0368fb77.jpg");"></li><li class="item" style="background-image: url("https://i.imgtg.com/2023/03/09/YS2LU.jpg");"></li><li class="item" style="background-image: url("https://img.timelessq.com/images/2022/07/26/9b626c5ba21d7cb4dbcba2b507688bbb.jpg");"></li></ul></div></header><div id="waves"><svg class="waves" xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" viewBox="0 24 150 28" preserveAspectRatio="none" shape-rendering="auto"><defs><path id="gentle-wave" d="M-160 44c30 0 58-18 88-18s 58 18 88 18 58-18 88-18 58 18 88 18 v44h-352z"></path></defs><g class="parallax"><use xlink:href="#gentle-wave" x="48" y="0"></use><use xlink:href="#gentle-wave" x="48" y="3"></use><use xlink:href="#gentle-wave" x="48" y="5"></use><use xlink:href="#gentle-wave" x="48" y="7"></use></g></svg></div><main><div class="inner"><div class="pjax" id="main"><div class="article wrap"><div class="breadcrumb" itemlistelement="" itemscope="" itemtype="https://schema.org/BreadcrumbList"><i class="ic i-home"></i><span><a href="/">首页</a></span><i class="ic i-angle-right"></i><span class="current" itemprop="itemListElement" itemscope="itemscope" itemtype="https://schema.org/ListItem"><a href="/categories/Coding/" itemprop="item" rel="index" title="分类于Coding"><span itemprop="name">Coding<meta itemprop="position" content="0"></span></a></span></div><article class="post block" itemscope="itemscope" itemtype="http://schema.org/Article" lang="zh-CN"><link itemprop="mainEntityOfPage" href="https://jiankychen.github.io/leetcode-hashtable.html"><span hidden="hidden" itemprop="author" itemscope="itemscope" itemtype="http://schema.org/Person"><meta itemprop="image" content="/assets/avatar.jpg"><meta itemprop="name" content="Jiankychen"><meta itemprop="description" content="Never put off till tomorrow what you can do today, "></span><span hidden="hidden" itemprop="publisher" itemscope="itemscope" itemtype="http://schema.org/Organization"><meta itemprop="name" content="Jiankychen's Blog"></span><div class="body md" itemprop="articleBody"><h1 id="leetcode-1-两数之和"><a class="anchor" href="#leetcode-1-两数之和">#</a> LeetCode 1. 两数之和</h1>
<p><a target="_blank" rel="noopener" href="https://leetcode-cn.com/problems/two-sum/">LeetCode 1. Two Sum</a></p>
<p>给定一个整数数组 <code>nums</code> 和一个整数目标值 <code>target</code> ，请你在该数组中找出 <strong>和为目标值</strong> <code>target</code> 的那 <strong>两个</strong> 整数，并返回它们的数组下标。</p>
<p>你可以假设每种输入只会对应一个答案。但是，数组中同一个元素在答案里不能重复出现。</p>
<p>你可以按任意顺序返回答案。</p>
<p></p>
<p><strong>示例 1：</strong></p>
<pre><code>输入：nums = [2,7,11,15], target = 9
输出：[0,1]
解释：因为 nums[0] + nums[1] == 9，返回 [0, 1]
</code></pre>
<p><strong>示例 2：</strong></p>
<pre><code>输入：nums = [3,2,4], target = 6
输出：[1,2]
</code></pre>
<p><strong>示例 3：</strong></p>
<pre><code>输入：nums = [3,3], target = 6
输出：[0,1]
</code></pre>
<p></p>
<p><strong>提示：</strong></p>
<ul>
<li><span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mn>2</mn></mrow><annotation encoding="application/x-tex">2</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.6444em;"></span><span class="mord">2</span></span></span></span> <= <code>nums.length</code> <= <span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mn>1</mn><msup><mn>0</mn><mn>4</mn></msup></mrow><annotation encoding="application/x-tex">10^4</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.8141em;"></span><span class="mord">1</span><span class="mord"><span class="mord">0</span><span class="msupsub"><span class="vlist-t"><span class="vlist-r"><span class="vlist" style="height:0.8141em;"><span style="top:-3.063em;margin-right:0.05em;"><span class="pstrut" style="height:2.7em;"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mtight">4</span></span></span></span></span></span></span></span></span></span></span></li>
<li><span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mo>−</mo><mn>1</mn><msup><mn>0</mn><mn>9</mn></msup></mrow><annotation encoding="application/x-tex">- 10^9</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.8974em;vertical-align:-0.0833em;"></span><span class="mord">−</span><span class="mord">1</span><span class="mord"><span class="mord">0</span><span class="msupsub"><span class="vlist-t"><span class="vlist-r"><span class="vlist" style="height:0.8141em;"><span style="top:-3.063em;margin-right:0.05em;"><span class="pstrut" style="height:2.7em;"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mtight">9</span></span></span></span></span></span></span></span></span></span></span> <= <code>nums[i]</code> <= <span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mn>1</mn><msup><mn>0</mn><mn>9</mn></msup></mrow><annotation encoding="application/x-tex">10^9</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.8141em;"></span><span class="mord">1</span><span class="mord"><span class="mord">0</span><span class="msupsub"><span class="vlist-t"><span class="vlist-r"><span class="vlist" style="height:0.8141em;"><span style="top:-3.063em;margin-right:0.05em;"><span class="pstrut" style="height:2.7em;"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mtight">9</span></span></span></span></span></span></span></span></span></span></span></li>
<li><span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mo>−</mo><mn>1</mn><msup><mn>0</mn><mn>9</mn></msup></mrow><annotation encoding="application/x-tex">- 10^9</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.8974em;vertical-align:-0.0833em;"></span><span class="mord">−</span><span class="mord">1</span><span class="mord"><span class="mord">0</span><span class="msupsub"><span class="vlist-t"><span class="vlist-r"><span class="vlist" style="height:0.8141em;"><span style="top:-3.063em;margin-right:0.05em;"><span class="pstrut" style="height:2.7em;"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mtight">9</span></span></span></span></span></span></span></span></span></span></span> <= <code>target</code> <= <span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mn>1</mn><msup><mn>0</mn><mn>9</mn></msup></mrow><annotation encoding="application/x-tex">10^9</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.8141em;"></span><span class="mord">1</span><span class="mord"><span class="mord">0</span><span class="msupsub"><span class="vlist-t"><span class="vlist-r"><span class="vlist" style="height:0.8141em;"><span style="top:-3.063em;margin-right:0.05em;"><span class="pstrut" style="height:2.7em;"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mtight">9</span></span></span></span></span></span></span></span></span></span></span></li>
<li>只存在一个有效答案</li>
</ul>
<p><strong>进阶</strong>：你可以想出一个时间复杂度小于 <span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>O</mi><mo stretchy="false">(</mo><msup><mi>n</mi><mn>2</mn></msup><mo stretchy="false">)</mo></mrow><annotation encoding="application/x-tex">O(n^2)</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:1.0641em;vertical-align:-0.25em;"></span><span class="mord mathnormal" style="margin-right:0.02778em;">O</span><span class="mopen">(</span><span class="mord"><span class="mord mathnormal">n</span><span class="msupsub"><span class="vlist-t"><span class="vlist-r"><span class="vlist" style="height:0.8141em;"><span style="top:-3.063em;margin-right:0.05em;"><span class="pstrut" style="height:2.7em;"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mtight">2</span></span></span></span></span></span></span></span><span class="mclose">)</span></span></span></span> 的算法吗？</p>
<h3 id="思路"><a class="anchor" href="#思路">#</a> 思路</h3>
<p>解题思路：遍历数组的下标 <code>i</code> ，判断 <code>target - nums[i]</code> 是否存在，若存在，返回 <code>i</code> 以及 数组中值为 <code>target - nums[i]</code> 的元素的下标</p>
<p>本题不仅要判读数值是否存在，还要记录其下标，因此采用哈希 map ，其中，元素值 作为 <code>key</code> ，元素在数组中的索引下标作为 <code>value</code></p>
<h2 id="method-哈希-map"><a class="anchor" href="#method-哈希-map">#</a> Method: 哈希 map</h2>
<p>由于并不需要 <code>key</code> 有序，可以选择 <code>unordered_map</code></p>
<p><a target="_blank" rel="noopener" href="http://www.cplusplus.com/reference/unordered_map/unordered_map/?kw=unordered_map">std::unoredered_map</a></p>
<p>解题步骤：</p>
<ol>
<li>
<p>定义 <code>unodered_map</code> 容器，命名为 <code>map</code></p>
</li>
<li>
<p>遍历数组下标 <code>i</code></p>
<ul>
<li>查找 <code>target - nums[i]</code> 是否存在于 <code>map</code> 当中</li>
<li>若存在，返回下标 <code>i</code> 以及 与键值 <code>target - nums[i]</code> 对应的 <code>value</code></li>
<li>否则，将 <code>nums[i]</code> 以及 <code>i</code> 作为键值对添加到 <code>map</code></li>
</ul>
</li>
</ol>
<p>代码实现：</p>
<figure class="highlight cpp"><figcaption data-lang="C++"></figcaption><table><tbody><tr><td data-num="1"></td><td><pre>vector<span class="token operator"><</span><span class="token keyword">int</span><span class="token operator">></span> <span class="token function">twoSum</span><span class="token punctuation">(</span>vector<span class="token operator"><</span><span class="token keyword">int</span><span class="token operator">></span><span class="token operator">&</span> nums<span class="token punctuation">,</span> <span class="token keyword">int</span> target<span class="token punctuation">)</span> <span class="token punctuation">{</span></pre></td></tr><tr><td data-num="2"></td><td><pre> unordered_map<span class="token operator"><</span><span class="token keyword">int</span><span class="token punctuation">,</span> <span class="token keyword">int</span><span class="token operator">></span> map<span class="token punctuation">;</span></pre></td></tr><tr><td data-num="3"></td><td><pre> <span class="token keyword">for</span> <span class="token punctuation">(</span><span class="token keyword">int</span> i <span class="token operator">=</span> <span class="token number">0</span><span class="token punctuation">;</span> i <span class="token operator"><</span> nums<span class="token punctuation">.</span><span class="token function">size</span><span class="token punctuation">(</span><span class="token punctuation">)</span><span class="token punctuation">;</span> i<span class="token operator">++</span><span class="token punctuation">)</span> <span class="token punctuation">{</span></pre></td></tr><tr><td data-num="4"></td><td><pre> <span class="token keyword">auto</span> temp <span class="token operator">=</span> map<span class="token punctuation">.</span><span class="token function">find</span><span class="token punctuation">(</span>target <span class="token operator">-</span> nums<span class="token punctuation">[</span>i<span class="token punctuation">]</span><span class="token punctuation">)</span><span class="token punctuation">;</span> <span class="token comment">//find 返回的是迭代器类型</span></pre></td></tr><tr><td data-num="5"></td><td><pre> <span class="token keyword">if</span> <span class="token punctuation">(</span>temp <span class="token operator">!=</span> map<span class="token punctuation">.</span><span class="token function">end</span><span class="token punctuation">(</span><span class="token punctuation">)</span><span class="token punctuation">)</span> <span class="token comment">// 找到 target - nums [i]</span></pre></td></tr><tr><td data-num="6"></td><td><pre> <span class="token keyword">return</span> <span class="token punctuation">{</span>temp<span class="token operator">-></span>second<span class="token punctuation">,</span> i<span class="token punctuation">}</span><span class="token punctuation">;</span> <span class="token comment">// 返回 i 以及 temp 的第二个值</span></pre></td></tr><tr><td data-num="7"></td><td><pre> <span class="token comment">//temp->first 是 key ，temp->second 是 value</span></pre></td></tr><tr><td data-num="8"></td><td><pre> map<span class="token punctuation">.</span><span class="token function">insert</span><span class="token punctuation">(</span><span class="token generic-function"><span class="token function">pair</span><span class="token generic class-name"><span class="token operator"><</span><span class="token keyword">int</span><span class="token punctuation">,</span> <span class="token keyword">int</span><span class="token operator">></span></span></span><span class="token punctuation">(</span>nums<span class="token punctuation">[</span>i<span class="token punctuation">]</span><span class="token punctuation">,</span> i<span class="token punctuation">)</span><span class="token punctuation">)</span><span class="token punctuation">;</span> <span class="token comment">// 将 nums [i], i 作为键值对插入到 map</span></pre></td></tr><tr><td data-num="9"></td><td><pre> <span class="token punctuation">}</span></pre></td></tr><tr><td data-num="10"></td><td><pre> <span class="token keyword">return</span> <span class="token punctuation">{</span><span class="token punctuation">}</span><span class="token punctuation">;</span></pre></td></tr><tr><td data-num="11"></td><td><pre><span class="token punctuation">}</span></pre></td></tr></tbody></table></figure><blockquote>
<p><code>unordered_map</code> 的成员函数 <code>find</code> 的返回值是 <code>iterator</code> 类型，解引用后的第一个成员为 <code>key</code> ，第二个成员为 <code>value</code> ，详见 <a target="_blank" rel="noopener" href="http://www.cplusplus.com/reference/unordered_map/unordered_map/find/">std::unordered_map::find</a></p>
</blockquote>
<h1 id="leetcode-128-最长连续序列"><a class="anchor" href="#leetcode-128-最长连续序列">#</a> LeetCode 128. 最长连续序列</h1>
<p><a target="_blank" rel="noopener" href="https://leetcode.cn/problems/longest-consecutive-sequence/">128. Longest Consecutive Sequence</a></p>
<p>给定一个未排序的整数数组 <code>nums</code> ，找出数字连续的最长序列（不要求序列元素在原数组中连续）的长度。</p>
<p>请你设计并实现时间复杂度为 <span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>O</mi><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow><annotation encoding="application/x-tex">O(n)</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:1em;vertical-align:-0.25em;"></span><span class="mord mathnormal" style="margin-right:0.02778em;">O</span><span class="mopen">(</span><span class="mord mathnormal">n</span><span class="mclose">)</span></span></span></span> 的算法解决此问题。</p>
<p><strong>示例 1：</strong></p>
<pre><code>输入：nums = [100,4,200,1,3,2]
输出：4
解释：The longest consecutive elements sequence is [1, 2, 3, 4]. Therefore its length is 4.
</code></pre>
<p><strong>示例 2：</strong></p>
<pre><code>输入：nums = [0,3,7,2,5,8,4,6,0,1]
输出：9
</code></pre>
<p><strong>提示：</strong></p>
<ul>
<li><span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mn>0</mn><mo>≤</mo></mrow><annotation encoding="application/x-tex">0 \le</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.7804em;vertical-align:-0.136em;"></span><span class="mord">0</span><span class="mspace" style="margin-right:0.2778em;"></span><span class="mrel">≤</span></span></span></span> <code>nums.length</code> <span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mo>≤</mo><mn>1</mn><msup><mn>0</mn><mn>5</mn></msup></mrow><annotation encoding="application/x-tex">\le 10^5</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.7719em;vertical-align:-0.136em;"></span><span class="mrel">≤</span><span class="mspace" style="margin-right:0.2778em;"></span></span><span class="base"><span class="strut" style="height:0.8141em;"></span><span class="mord">1</span><span class="mord"><span class="mord">0</span><span class="msupsub"><span class="vlist-t"><span class="vlist-r"><span class="vlist" style="height:0.8141em;"><span style="top:-3.063em;margin-right:0.05em;"><span class="pstrut" style="height:2.7em;"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mtight">5</span></span></span></span></span></span></span></span></span></span></span></li>
<li><span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mo>−</mo><mn>1</mn><msup><mn>0</mn><mn>9</mn></msup><mo>≤</mo></mrow><annotation encoding="application/x-tex">-10^9 \le</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.9501em;vertical-align:-0.136em;"></span><span class="mord">−</span><span class="mord">1</span><span class="mord"><span class="mord">0</span><span class="msupsub"><span class="vlist-t"><span class="vlist-r"><span class="vlist" style="height:0.8141em;"><span style="top:-3.063em;margin-right:0.05em;"><span class="pstrut" style="height:2.7em;"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mtight">9</span></span></span></span></span></span></span></span><span class="mspace" style="margin-right:0.2778em;"></span><span class="mrel">≤</span></span></span></span> <code>nums[i]</code> <span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mo>≤</mo><mn>1</mn><msup><mn>0</mn><mn>9</mn></msup></mrow><annotation encoding="application/x-tex">\le 10^9</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.7719em;vertical-align:-0.136em;"></span><span class="mrel">≤</span><span class="mspace" style="margin-right:0.2778em;"></span></span><span class="base"><span class="strut" style="height:0.8141em;"></span><span class="mord">1</span><span class="mord"><span class="mord">0</span><span class="msupsub"><span class="vlist-t"><span class="vlist-r"><span class="vlist" style="height:0.8141em;"><span style="top:-3.063em;margin-right:0.05em;"><span class="pstrut" style="height:2.7em;"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mtight">9</span></span></span></span></span></span></span></span></span></span></span></li>
</ul>
<h2 id="method-哈希表"><a class="anchor" href="#method-哈希表">#</a> Method: 哈希表</h2>
<p>算法思路：</p>
<p>定义一个哈希表（记作 hash ，即 <code>unordered_map<int> hash</code> ），将数组中的数字作为哈希表的键，将数字所在的连续序列的最大长度作为对应的哈希表值</p>
<p>遍历数组（记当前遍历到的数字为 num）：</p>
<ul>
<li>
<p>若数字 num 已在哈希表中，跳过</p>
</li>
<li>
<p>若数字 num 不在哈希表中：</p>
<ul>
<li>
<p>找出左侧相邻数 num - 1 所在的连续序列的最大长度，记作 left ，即 left = hash [num - 1]</p>
</li>
<li>
<p>找出右侧相邻数 num + 1 所在的连续序列的最大长度，记作 right ，即 right = hash [num + 1]</p>
</li>
<li>
<p>将 num 左右两侧的连续序列以及 num 进行拼接，计算新的连续序列的长度：len = left + right + 1</p>
</li>
<li>
<p>更新当前数字 num 对应的哈希表值，即 hash [num] = len</p>
</li>
<li>
<p>更新连续序列两个端点对应的哈希表值，即，hash [num - left] = len ，hash [num + right] = len</p>
</li>
<li>
<p>更新整个数组中的最长连续序列的长度，即 ans = max (ans, len)</p>
</li>
</ul>
</li>
</ul>
<p>上述算法中，拼接 num 及其左右两侧连续序列时，我们没有更新连续序列中每个数字对应的哈希表值，而只是更新两个端点对应的哈希表值。这是因为：区间 (num - left, num + right) 内数字对应的哈希表值不会被使用，只有 num -left 和 num + right 这两个端点对应的哈希表值可能被使用</p>
<ul>
<li>在之后的遍历中，如果遇到 [num - left, num + right] 区间内的数，会直接跳过，无需考察其左右两侧数字所在连续序列的长度</li>
<li>在之后的遍历中，如果遇到 num - left - 1（或 num + right + 1），由于 num - left（或 num + right）对应的连续序列长度已经更新，可以直接将 num - left - 1（或 num + right + 1）与 num - left（或 num + right）对应的连续序列进行拼接，无需担心结果出错</li>
</ul>
<blockquote>
<p>无论 num 左右两侧是否存在相应的连续序列，上述算法都能奏效（哈希表的值全都初始化为 0 ）</p>
<ul>
<li>当 num 左侧或右侧不存在连续序列时，left 或 right 为 0，num - left 或 num + right 就是 num 本身</li>
<li>当 num 左右两侧均不存在连续序列时，left 和 right 均为 0，num - left 和 num + right 都是 num 本身</li>
</ul>
</blockquote>
<p>代码实现：</p>
<figure class="highlight cpp"><figcaption data-lang="C++"></figcaption><table><tbody><tr><td data-num="1"></td><td><pre><span class="token keyword">int</span> <span class="token function">longestConsecutive</span><span class="token punctuation">(</span>vector<span class="token operator"><</span><span class="token keyword">int</span><span class="token operator">></span><span class="token operator">&</span> nums<span class="token punctuation">)</span> <span class="token punctuation">{</span></pre></td></tr><tr><td data-num="2"></td><td><pre> unordered_map<span class="token operator"><</span><span class="token keyword">int</span><span class="token punctuation">,</span> <span class="token keyword">int</span><span class="token operator">></span> hash<span class="token punctuation">;</span> <span class="token comment">// 哈希表，存储每个数字对应连续序列的长度</span></pre></td></tr><tr><td data-num="3"></td><td><pre> <span class="token keyword">int</span> left <span class="token operator">=</span> <span class="token number">0</span><span class="token punctuation">;</span></pre></td></tr><tr><td data-num="4"></td><td><pre> <span class="token keyword">int</span> right <span class="token operator">=</span> <span class="token number">0</span><span class="token punctuation">;</span></pre></td></tr><tr><td data-num="5"></td><td><pre> <span class="token keyword">int</span> len <span class="token operator">=</span> <span class="token number">0</span><span class="token punctuation">;</span></pre></td></tr><tr><td data-num="6"></td><td><pre> <span class="token keyword">int</span> ans <span class="token operator">=</span> <span class="token number">0</span><span class="token punctuation">;</span></pre></td></tr><tr><td data-num="7"></td><td><pre> <span class="token keyword">for</span> <span class="token punctuation">(</span><span class="token keyword">int</span> num <span class="token operator">:</span> nums<span class="token punctuation">)</span> <span class="token punctuation">{</span></pre></td></tr><tr><td data-num="8"></td><td><pre> <span class="token keyword">if</span> <span class="token punctuation">(</span>hash<span class="token punctuation">[</span>num<span class="token punctuation">]</span> <span class="token operator">==</span> <span class="token number">0</span><span class="token punctuation">)</span> <span class="token punctuation">{</span></pre></td></tr><tr><td data-num="9"></td><td><pre> left <span class="token operator">=</span> hash<span class="token punctuation">[</span>num <span class="token operator">-</span> <span class="token number">1</span><span class="token punctuation">]</span><span class="token punctuation">;</span> <span class="token comment">// 左侧数字对应连续序列的长度</span></pre></td></tr><tr><td data-num="10"></td><td><pre> right <span class="token operator">=</span> hash<span class="token punctuation">[</span>num <span class="token operator">+</span> <span class="token number">1</span><span class="token punctuation">]</span><span class="token punctuation">;</span> <span class="token comment">// 右侧数字对应连续序列的长度</span></pre></td></tr><tr><td data-num="11"></td><td><pre> len <span class="token operator">=</span> left <span class="token operator">+</span> right <span class="token operator">+</span> <span class="token number">1</span><span class="token punctuation">;</span> <span class="token comment">// 拼接左右连续序列</span></pre></td></tr><tr><td data-num="12"></td><td><pre> hash<span class="token punctuation">[</span>num<span class="token punctuation">]</span> <span class="token operator">=</span> len<span class="token punctuation">;</span> <span class="token comment">// 数字 num 对应连续序列的长度</span></pre></td></tr><tr><td data-num="13"></td><td><pre> hash<span class="token punctuation">[</span>num <span class="token operator">-</span> left<span class="token punctuation">]</span> <span class="token operator">=</span> len<span class="token punctuation">;</span> <span class="token comment">// 更新左侧数字对应序列的端点的键值</span></pre></td></tr><tr><td data-num="14"></td><td><pre> hash<span class="token punctuation">[</span>num <span class="token operator">+</span> right<span class="token punctuation">]</span> <span class="token operator">=</span> len<span class="token punctuation">;</span> <span class="token comment">// 更新右侧数字对应序列的端点的键值 </span></pre></td></tr><tr><td data-num="15"></td><td><pre> <span class="token keyword">if</span> <span class="token punctuation">(</span>len <span class="token operator">></span> ans<span class="token punctuation">)</span> ans <span class="token operator">=</span> len<span class="token punctuation">;</span> <span class="token comment">// 更新最大长度 </span></pre></td></tr><tr><td data-num="16"></td><td><pre> <span class="token punctuation">}</span></pre></td></tr><tr><td data-num="17"></td><td><pre> <span class="token punctuation">}</span></pre></td></tr><tr><td data-num="18"></td><td><pre> <span class="token keyword">return</span> ans<span class="token punctuation">;</span></pre></td></tr><tr><td data-num="19"></td><td><pre><span class="token punctuation">}</span></pre></td></tr></tbody></table></figure><p>时间复杂度：<span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>O</mi><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow><annotation encoding="application/x-tex">O(n)</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:1em;vertical-align:-0.25em;"></span><span class="mord mathnormal" style="margin-right:0.02778em;">O</span><span class="mopen">(</span><span class="mord mathnormal">n</span><span class="mclose">)</span></span></span></span>，其中 <span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>n</mi></mrow><annotation encoding="application/x-tex">n</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.4306em;"></span><span class="mord mathnormal">n</span></span></span></span> 是数组的长度</p>
<p>空间复杂度：<span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>O</mi><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow><annotation encoding="application/x-tex">O(n)</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:1em;vertical-align:-0.25em;"></span><span class="mord mathnormal" style="margin-right:0.02778em;">O</span><span class="mopen">(</span><span class="mord mathnormal">n</span><span class="mclose">)</span></span></span></span></p>
<p>参考：<a target="_blank" rel="noopener" href="https://leetcode.cn/problems/longest-consecutive-sequence/solution/dong-tai-gui-hua-python-ti-jie-by-jalan/">jalan</a></p>
<h1 id="leetcode-1296-划分数组为连续数字的集合"><a class="anchor" href="#leetcode-1296-划分数组为连续数字的集合">#</a> LeetCode 1296. 划分数组为连续数字的集合</h1>
<p><a target="_blank" rel="noopener" href="https://leetcode.cn/problems/divide-array-in-sets-of-k-consecutive-numbers/description/">1296. 划分数组为连续数字的集合</a></p>
<p>给你一个整数数组 <code>nums</code> 和一个正整数 <code>k</code> ，请你判断是否可以把这个数组划分成一些由 <code>k</code> 个连续数字组成的集合</p>
<p>如果可以，请返回 <code>true</code> ；否则，返回 <code>false</code></p>
<p><strong>示例 1：</strong></p>
<pre><code>输入：nums = [1,2,3,3,4,4,5,6], k = 4
输出：true
解释：数组可以分成 [1,2,3,4] 和 [3,4,5,6]
</code></pre>
<p><strong>示例 2：</strong></p>
<pre><code>输入：nums = [3,2,1,2,3,4,3,4,5,9,10,11], k = 3
输出：true
解释：数组可以分成 [1,2,3] , [2,3,4] , [3,4,5] 和 [9,10,11]
</code></pre>
<p><strong>示例 3：</strong></p>
<pre><code>输入：nums = [3,3,2,2,1,1], k = 3
输出：true
</code></pre>
<p><strong>示例 4：</strong></p>
<pre><code>输入：nums = [1,2,3,4], k = 3
输出：false
解释：数组不能分成几个大小为 3 的子数组。
</code></pre>
<p><strong>提示：</strong></p>
<ul>
<li><span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mn>1</mn><mo>≤</mo></mrow><annotation encoding="application/x-tex">1 \le</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.7804em;vertical-align:-0.136em;"></span><span class="mord">1</span><span class="mspace" style="margin-right:0.2778em;"></span><span class="mrel">≤</span></span></span></span> <code>k</code> <span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mo>≤</mo></mrow><annotation encoding="application/x-tex">\le</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.7719em;vertical-align:-0.136em;"></span><span class="mrel">≤</span></span></span></span> <code>nums.length</code> <span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mo>≤</mo><mn>1</mn><msup><mn>0</mn><mn>5</mn></msup></mrow><annotation encoding="application/x-tex">\le 10^5</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.7719em;vertical-align:-0.136em;"></span><span class="mrel">≤</span><span class="mspace" style="margin-right:0.2778em;"></span></span><span class="base"><span class="strut" style="height:0.8141em;"></span><span class="mord">1</span><span class="mord"><span class="mord">0</span><span class="msupsub"><span class="vlist-t"><span class="vlist-r"><span class="vlist" style="height:0.8141em;"><span style="top:-3.063em;margin-right:0.05em;"><span class="pstrut" style="height:2.7em;"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mtight">5</span></span></span></span></span></span></span></span></span></span></span></li>
<li><span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mn>1</mn><mo>≤</mo></mrow><annotation encoding="application/x-tex">1 \le</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.7804em;vertical-align:-0.136em;"></span><span class="mord">1</span><span class="mspace" style="margin-right:0.2778em;"></span><span class="mrel">≤</span></span></span></span> <code>nums[i]</code> <span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mo>≤</mo><mn>1</mn><msup><mn>0</mn><mn>9</mn></msup></mrow><annotation encoding="application/x-tex">\le 10^9</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.7719em;vertical-align:-0.136em;"></span><span class="mrel">≤</span><span class="mspace" style="margin-right:0.2778em;"></span></span><span class="base"><span class="strut" style="height:0.8141em;"></span><span class="mord">1</span><span class="mord"><span class="mord">0</span><span class="msupsub"><span class="vlist-t"><span class="vlist-r"><span class="vlist" style="height:0.8141em;"><span style="top:-3.063em;margin-right:0.05em;"><span class="pstrut" style="height:2.7em;"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mtight">9</span></span></span></span></span></span></span></span></span></span></span></li>
</ul>
<h2 id="method-排序-哈希"><a class="anchor" href="#method-排序-哈希">#</a> Method: 排序 + 哈希</h2>
<h2 id="算法思路"><a class="anchor" href="#算法思路">#</a> 算法思路</h2>
<p>题目要求将数组划分成若干个集合，其中，每个集合包含 k 个连续数字</p>
<p>可以从尚未分组的元素中找出值最小的元素，将其作为集合的第一个元素（记作 x ），于是该集合中数字的范围应为 [x, x + k - 1] 。如果某个数字不存在，则无法将数组划分成符合条件的集合，返回 false</p>
<p>将 [x, x + k - 1] 这 k 个元素划分到一个集合之后，继续对数组中剩余的数字进行分组，直到 所有元素均已分组 或者 遇到无法分组的情况</p>
<h2 id="代码实现"><a class="anchor" href="#代码实现">#</a> 代码实现</h2>
<figure class="highlight cpp"><figcaption data-lang="C++"></figcaption><table><tbody><tr><td data-num="1"></td><td><pre><span class="token keyword">bool</span> <span class="token function">isPossibleDivide</span><span class="token punctuation">(</span>vector<span class="token operator"><</span><span class="token keyword">int</span><span class="token operator">></span><span class="token operator">&</span> nums<span class="token punctuation">,</span> <span class="token keyword">int</span> k<span class="token punctuation">)</span> <span class="token punctuation">{</span></pre></td></tr><tr><td data-num="2"></td><td><pre> <span class="token keyword">if</span> <span class="token punctuation">(</span>nums<span class="token punctuation">.</span><span class="token function">size</span><span class="token punctuation">(</span><span class="token punctuation">)</span> <span class="token operator">%</span> k<span class="token punctuation">)</span> <span class="token keyword">return</span> <span class="token boolean">false</span><span class="token punctuation">;</span> <span class="token comment">// 数组长度无法被 k 整除，返回 false</span></pre></td></tr><tr><td data-num="3"></td><td><pre> <span class="token function">sort</span><span class="token punctuation">(</span>nums<span class="token punctuation">.</span><span class="token function">begin</span><span class="token punctuation">(</span><span class="token punctuation">)</span><span class="token punctuation">,</span> nums<span class="token punctuation">.</span><span class="token function">end</span><span class="token punctuation">(</span><span class="token punctuation">)</span><span class="token punctuation">)</span><span class="token punctuation">;</span> <span class="token comment">// 将数组按从小到大排序</span></pre></td></tr><tr><td data-num="4"></td><td><pre> unordered_map<span class="token operator"><</span><span class="token keyword">int</span><span class="token punctuation">,</span> <span class="token keyword">int</span><span class="token operator">></span> hashmap<span class="token punctuation">;</span></pre></td></tr><tr><td data-num="5"></td><td><pre> <span class="token keyword">for</span> <span class="token punctuation">(</span><span class="token keyword">auto</span> num <span class="token operator">:</span> nums<span class="token punctuation">)</span> <span class="token operator">++</span>hashmap<span class="token punctuation">[</span>num<span class="token punctuation">]</span><span class="token punctuation">;</span> <span class="token comment">// 统计每个数字的出现次数</span></pre></td></tr><tr><td data-num="6"></td><td><pre> <span class="token keyword">for</span> <span class="token punctuation">(</span><span class="token keyword">int</span> i <span class="token operator">=</span> <span class="token number">0</span><span class="token punctuation">;</span> i <span class="token operator"><</span> nums<span class="token punctuation">.</span><span class="token function">size</span><span class="token punctuation">(</span><span class="token punctuation">)</span> <span class="token operator">/</span> k<span class="token punctuation">;</span> <span class="token operator">++</span>i<span class="token punctuation">)</span> <span class="token punctuation">{</span> <span class="token comment">// 一共有 nums.size () /k 个集合</span></pre></td></tr><tr><td data-num="7"></td><td><pre> <span class="token keyword">int</span> first <span class="token operator">=</span> <span class="token number">0</span><span class="token punctuation">;</span> <span class="token comment">// 集合中第一个元素的下标</span></pre></td></tr><tr><td data-num="8"></td><td><pre> <span class="token keyword">while</span> <span class="token punctuation">(</span>first <span class="token operator"><</span> nums<span class="token punctuation">.</span><span class="token function">size</span><span class="token punctuation">(</span><span class="token punctuation">)</span> <span class="token operator">&&</span> hashmap<span class="token punctuation">[</span>nums<span class="token punctuation">[</span>first<span class="token punctuation">]</span><span class="token punctuation">]</span> <span class="token operator">==</span> <span class="token number">0</span><span class="token punctuation">)</span> <span class="token punctuation">{</span> <span class="token comment">// 寻找未被使用的、值最小的元素</span></pre></td></tr><tr><td data-num="9"></td><td><pre> <span class="token operator">++</span>first<span class="token punctuation">;</span></pre></td></tr><tr><td data-num="10"></td><td><pre> <span class="token punctuation">}</span></pre></td></tr><tr><td data-num="11"></td><td><pre> <span class="token operator">--</span>hashmap<span class="token punctuation">[</span>nums<span class="token punctuation">[</span>first<span class="token punctuation">]</span><span class="token punctuation">]</span><span class="token punctuation">;</span> <span class="token comment">// 将 nums [first] 添加到集合（可用次数减 1 ）</span></pre></td></tr><tr><td data-num="12"></td><td><pre> <span class="token keyword">for</span> <span class="token punctuation">(</span><span class="token keyword">int</span> j <span class="token operator">=</span> <span class="token number">1</span><span class="token punctuation">;</span> j <span class="token operator"><</span> k<span class="token punctuation">;</span> <span class="token operator">++</span>j<span class="token punctuation">)</span> <span class="token punctuation">{</span> <span class="token comment">// 寻找剩余的 k - 1 个元素</span></pre></td></tr><tr><td data-num="13"></td><td><pre> <span class="token keyword">if</span> <span class="token punctuation">(</span>hashmap<span class="token punctuation">[</span>nums<span class="token punctuation">[</span>first<span class="token punctuation">]</span> <span class="token operator">+</span> j<span class="token punctuation">]</span> <span class="token operator">==</span> <span class="token number">0</span><span class="token punctuation">)</span> <span class="token comment">// 不存在 nums [first] + j 这个数，数字不连续，返回 false</span></pre></td></tr><tr><td data-num="14"></td><td><pre> <span class="token keyword">return</span> <span class="token boolean">false</span><span class="token punctuation">;</span></pre></td></tr><tr><td data-num="15"></td><td><pre> <span class="token keyword">else</span> <span class="token comment">// 存在 nums [first] + j 这个数，将其可用次数减 1</span></pre></td></tr><tr><td data-num="16"></td><td><pre> <span class="token operator">--</span>hashmap<span class="token punctuation">[</span>nums<span class="token punctuation">[</span>first<span class="token punctuation">]</span> <span class="token operator">+</span> j<span class="token punctuation">]</span><span class="token punctuation">;</span></pre></td></tr><tr><td data-num="17"></td><td><pre> <span class="token punctuation">}</span></pre></td></tr><tr><td data-num="18"></td><td><pre> <span class="token punctuation">}</span></pre></td></tr><tr><td data-num="19"></td><td><pre> <span class="token keyword">return</span> <span class="token boolean">true</span><span class="token punctuation">;</span></pre></td></tr><tr><td data-num="20"></td><td><pre><span class="token punctuation">}</span></pre></td></tr></tbody></table></figure><h2 id="复杂度分析"><a class="anchor" href="#复杂度分析">#</a> 复杂度分析</h2>
<p>时间复杂度：<span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>O</mi><mo stretchy="false">(</mo><mi>n</mi><mi>log</mi><mo></mo><mi>n</mi><mo stretchy="false">)</mo></mrow><annotation encoding="application/x-tex">O(n \log{n})</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:1em;vertical-align:-0.25em;"></span><span class="mord mathnormal" style="margin-right:0.02778em;">O</span><span class="mopen">(</span><span class="mord mathnormal">n</span><span class="mspace" style="margin-right:0.1667em;"></span><span class="mop">lo<span style="margin-right:0.01389em;">g</span></span><span class="mspace" style="margin-right:0.1667em;"></span><span class="mord"><span class="mord mathnormal">n</span></span><span class="mclose">)</span></span></span></span>，其中，<span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>n</mi></mrow><annotation encoding="application/x-tex">n</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.4306em;"></span><span class="mord mathnormal">n</span></span></span></span> 为数组的长度</p>
<p>空间复杂度：<span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>O</mi><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow><annotation encoding="application/x-tex">O(n)</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:1em;vertical-align:-0.25em;"></span><span class="mord mathnormal" style="margin-right:0.02778em;">O</span><span class="mopen">(</span><span class="mord mathnormal">n</span><span class="mclose">)</span></span></span></span>，考虑哈希表所需空间</p>
<h2 id="参考资料"><a class="anchor" href="#参考资料">#</a> 参考资料</h2>
<p>参考：<a target="_blank" rel="noopener" href="https://leetcode.cn/problems/divide-array-in-sets-of-k-consecutive-numbers/solutions/101809/hua-fen-shu-zu-wei-lian-xu-shu-zi-de-ji-he-by-le-2/">力扣官方题解</a></p>
<h1 id="leetcode-15-三数之和"><a class="anchor" href="#leetcode-15-三数之和">#</a> LeetCode 15. 三数之和</h1>
<p><a target="_blank" rel="noopener" href="https://leetcode-cn.com/problems/3sum/">LeetCode 15. Three Sum</a></p>
<p>给你一个整数数组 <code>nums</code> ，判断是否存在三元组 <code>[nums[i]</code> , <code>nums[j]</code> , <code>nums[k]]</code> 满足 <code>i != j</code> 、 <code>i != k</code> 且 <code>j != k</code> ，同时还满足 <code>nums[i] + nums[j] + nums[k] == 0</code> 。请你返回所有和为 0 且不重复的三元组。</p>
<p><strong>注意</strong>：答案中不可以包含重复的三元组。</p>
<p></p>
<p><strong>示例 1：</strong></p>
<pre><code>输入：nums = [-1,0,1,2,-1,-4]
输出：[[-1,-1,2],[-1,0,1]]
解释：
nums[0] + nums[1] + nums[2] = (-1) + 0 + 1 = 0 。
nums[1] + nums[2] + nums[4] = 0 + 1 + (-1) = 0 。
nums[0] + nums[3] + nums[4] = (-1) + 2 + (-1) = 0 。
不同的三元组是 [-1,0,1] 和 [-1,-1,2] 。
注意，输出的顺序和三元组的顺序并不重要。
</code></pre>
<p><strong>示例 2：</strong></p>
<pre><code>输入：nums = []
输出：[]
解释：唯一可能的三元组和不为 0 。
</code></pre>
<p><strong>示例 3：</strong></p>
<pre><code>输入：nums = [0]
输出：[]
解释：唯一可能的三元组和为 0 。
</code></pre>
<p></p>
<p><strong>提示：</strong></p>
<ul>
<li><span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mn>0</mn><mo>≤</mo></mrow><annotation encoding="application/x-tex">0 \le</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.7804em;vertical-align:-0.136em;"></span><span class="mord">0</span><span class="mspace" style="margin-right:0.2778em;"></span><span class="mrel">≤</span></span></span></span> <code>nums.length</code> <span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mo>≤</mo><mn>3000</mn></mrow><annotation encoding="application/x-tex">\le 3000</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.7719em;vertical-align:-0.136em;"></span><span class="mrel">≤</span><span class="mspace" style="margin-right:0.2778em;"></span></span><span class="base"><span class="strut" style="height:0.6444em;"></span><span class="mord">3000</span></span></span></span></li>
<li><span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mo>−</mo><mn>1</mn><msup><mn>0</mn><mn>5</mn></msup><mo>≤</mo></mrow><annotation encoding="application/x-tex">- 10^5 \le</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.9501em;vertical-align:-0.136em;"></span><span class="mord">−</span><span class="mord">1</span><span class="mord"><span class="mord">0</span><span class="msupsub"><span class="vlist-t"><span class="vlist-r"><span class="vlist" style="height:0.8141em;"><span style="top:-3.063em;margin-right:0.05em;"><span class="pstrut" style="height:2.7em;"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mtight">5</span></span></span></span></span></span></span></span><span class="mspace" style="margin-right:0.2778em;"></span><span class="mrel">≤</span></span></span></span> <code>nums[i]</code> <span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mo>≤</mo><mn>1</mn><msup><mn>0</mn><mn>5</mn></msup></mrow><annotation encoding="application/x-tex">\le 10^5</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.7719em;vertical-align:-0.136em;"></span><span class="mrel">≤</span><span class="mspace" style="margin-right:0.2778em;"></span></span><span class="base"><span class="strut" style="height:0.8141em;"></span><span class="mord">1</span><span class="mord"><span class="mord">0</span><span class="msupsub"><span class="vlist-t"><span class="vlist-r"><span class="vlist" style="height:0.8141em;"><span style="top:-3.063em;margin-right:0.05em;"><span class="pstrut" style="height:2.7em;"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mtight">5</span></span></span></span></span></span></span></span></span></span></span></li>
</ul>
<h2 id="思路-2"><a class="anchor" href="#思路-2">#</a> 思路</h2>
<p>暴力法查找的时间复杂度为 <span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>O</mi><mo stretchy="false">(</mo><msup><mi>n</mi><mn>3</mn></msup><mo stretchy="false">)</mo></mrow><annotation encoding="application/x-tex">O(n^3)</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:1.0641em;vertical-align:-0.25em;"></span><span class="mord mathnormal" style="margin-right:0.02778em;">O</span><span class="mopen">(</span><span class="mord"><span class="mord mathnormal">n</span><span class="msupsub"><span class="vlist-t"><span class="vlist-r"><span class="vlist" style="height:0.8141em;"><span style="top:-3.063em;margin-right:0.05em;"><span class="pstrut" style="height:2.7em;"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mtight">3</span></span></span></span></span></span></span></span><span class="mclose">)</span></span></span></span> ，考虑用低复杂度算法</p>
<p>注意：题目要求 “ <code>i != j</code> , <code>i != k</code> , <code>j != k</code> ” 并且 “任意两个三元组不能相同” ，故而需要在查找时进行剪枝以避免重复，或是查找结束后剔除重复三元组</p>
<p>本题的难点也在于如何去除重复解</p>
<p>首先，可以对数组进行排序（最终只需要输出元素值即可，无需输出元素索引，故而可以打乱原数组顺序），将重复的元素值集中，便于去重</p>
<p>于是，有以下两种方法可以用于查找：</p>
<ol>
<li>
<p>哈希法：采用两层 <code>for</code> 循环分别遍历 <code>i</code> 和 <code>j</code> （按照从左往右的顺序），并采用哈希法检查 <code>[i, j]</code> 区间范围内是否有元素能与 <code>nums[i]</code> ， <code>nums[j]</code> 组成三元组</p>
</li>
<li>
<p>双指针法：采用一个 <code>for</code> 循环遍历 <code>i</code> ，并采用双指针法在 <code>[i + 1, nums.size() - 1]</code> 区间内查找所有能与 <code>nums[i]</code> 组成三元组的元素 <code>nums[left]</code> 和 <code>nums[right]</code> ，其中，指针 <code>left</code> 从 <code>i + 1</code> 位置开始向右遍历，指针 <code>right</code> 从 <code>nums.size() - 1</code> 位置开始向左遍历</p>
</li>
</ol>
<p>两种方法都能实现 <span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>O</mi><mo stretchy="false">(</mo><msup><mi>n</mi><mn>2</mn></msup><mo stretchy="false">)</mo></mrow><annotation encoding="application/x-tex">O(n^2)</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:1.0641em;vertical-align:-0.25em;"></span><span class="mord mathnormal" style="margin-right:0.02778em;">O</span><span class="mopen">(</span><span class="mord"><span class="mord mathnormal">n</span><span class="msupsub"><span class="vlist-t"><span class="vlist-r"><span class="vlist" style="height:0.8141em;"><span style="top:-3.063em;margin-right:0.05em;"><span class="pstrut" style="height:2.7em;"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mtight">2</span></span></span></span></span></span></span></span><span class="mclose">)</span></span></span></span> 的时间复杂度，但哈希法不便进行去重操作，因此，建议使用排序与双指针法解题</p>
<h2 id="method-排序-双指针"><a class="anchor" href="#method-排序-双指针">#</a> Method: 排序 + 双指针</h2>
<p>解题步骤：</p>
<ol>
<li>
<p>对数组进行排序（从小到大排序）</p>
</li>
<li>
<p>遍历数组下标 <code>i</code></p>
<ul>
<li>
<p>若 <code>nums[i] > 0</code> ，则 <code>i</code> 右侧不存在能与 <code>nums[i]</code> 组成三元组的元素，直接返回结果</p>
</li>
<li>
<p>若 <code>i > 0 && nums[i] == nums[i - 1]</code> ，当前 <code>i</code> 能找到的三元组与 <code>i - 1</code> 时找到的完全相同，为避免产生重复解，跳过当前 <code>i</code></p>
</li>
<li>
<p>定义指针 <code>left</code> 指向 <code>i + 1</code> 位置，指针 <code>right</code> 指向 <code>nums.size() - 1</code> 位置，当 <code>left < right</code> 时，执行循环：</p>
<ul>
<li>计算 <code>sum = nums[i] + nums[left] + nums[right]</code></li>
<li>若 <code>sum > 0</code> ，则 <code>nums[right]</code> 偏大，将 <code>right</code> 左移</li>
<li>若 <code>sum < 0</code> ，则 <code>nums[left]</code> 偏小，将 <code>left</code> 右移</li>
<li>若 <code>sum == 0</code> ，记录结果，并将 <code>left</code> 右移、将 <code>right</code> 左移，以跳过重复的 <code>nums[left]</code> 和 <code>nums[right]</code></li>
</ul>
</li>
</ul>
</li>
</ol>
<blockquote>
<p>注意：同一个 <code>i</code> 可以与不同的元素组成多个不同的三元组，因此，在找到一对可行的 <code>nums[left]</code> 和 <code>nums[right]</code> 后仍需继续查找，直到 <code>left < right</code> 不满足</p>
</blockquote>
<p>代码实现：</p>
<figure class="highlight cpp"><figcaption data-lang="C++"></figcaption><table><tbody><tr><td data-num="1"></td><td><pre>vector<span class="token operator"><</span>vector<span class="token operator"><</span><span class="token keyword">int</span><span class="token operator">>></span> <span class="token function">threeSum</span><span class="token punctuation">(</span>vector<span class="token operator"><</span><span class="token keyword">int</span><span class="token operator">></span><span class="token operator">&</span> nums<span class="token punctuation">)</span> <span class="token punctuation">{</span></pre></td></tr><tr><td data-num="2"></td><td><pre> <span class="token function">sort</span><span class="token punctuation">(</span>nums<span class="token punctuation">.</span><span class="token function">begin</span><span class="token punctuation">(</span><span class="token punctuation">)</span><span class="token punctuation">,</span> nums<span class="token punctuation">.</span><span class="token function">end</span><span class="token punctuation">(</span><span class="token punctuation">)</span><span class="token punctuation">)</span><span class="token punctuation">;</span> <span class="token comment">// 排序</span></pre></td></tr><tr><td data-num="3"></td><td><pre> vector<span class="token operator"><</span>vector<span class="token operator"><</span><span class="token keyword">int</span><span class="token operator">>></span> res<span class="token punctuation">;</span> <span class="token comment">// 存储结果</span></pre></td></tr><tr><td data-num="4"></td><td><pre> <span class="token keyword">for</span> <span class="token punctuation">(</span><span class="token keyword">int</span> i <span class="token operator">=</span> <span class="token number">0</span><span class="token punctuation">;</span> i <span class="token operator"><</span> nums<span class="token punctuation">.</span><span class="token function">size</span><span class="token punctuation">(</span><span class="token punctuation">)</span><span class="token punctuation">;</span> i<span class="token operator">++</span><span class="token punctuation">)</span> <span class="token punctuation">{</span></pre></td></tr><tr><td data-num="5"></td><td><pre> <span class="token keyword">if</span> <span class="token punctuation">(</span>nums<span class="token punctuation">[</span>i<span class="token punctuation">]</span> <span class="token operator">></span> <span class="token number">0</span><span class="token punctuation">)</span> <span class="token keyword">break</span><span class="token punctuation">;</span></pre></td></tr><tr><td data-num="6"></td><td><pre> <span class="token keyword">if</span> <span class="token punctuation">(</span>i <span class="token operator">></span> <span class="token number">0</span> <span class="token operator">&&</span> nums<span class="token punctuation">[</span>i<span class="token punctuation">]</span> <span class="token operator">==</span> nums<span class="token punctuation">[</span>i <span class="token operator">-</span> <span class="token number">1</span><span class="token punctuation">]</span><span class="token punctuation">)</span> <span class="token keyword">continue</span><span class="token punctuation">;</span></pre></td></tr><tr><td data-num="7"></td><td><pre> <span class="token keyword">int</span> left <span class="token operator">=</span> i <span class="token operator">+</span> <span class="token number">1</span><span class="token punctuation">,</span> right <span class="token operator">=</span> nums<span class="token punctuation">.</span><span class="token function">size</span><span class="token punctuation">(</span><span class="token punctuation">)</span> <span class="token operator">-</span> <span class="token number">1</span><span class="token punctuation">;</span></pre></td></tr><tr><td data-num="8"></td><td><pre> <span class="token keyword">int</span> sum <span class="token operator">=</span> <span class="token number">0</span><span class="token punctuation">;</span></pre></td></tr><tr><td data-num="9"></td><td><pre> <span class="token keyword">while</span> <span class="token punctuation">(</span>left <span class="token operator"><</span> right<span class="token punctuation">)</span> <span class="token punctuation">{</span></pre></td></tr><tr><td data-num="10"></td><td><pre> sum <span class="token operator">=</span> nums<span class="token punctuation">[</span>i<span class="token punctuation">]</span> <span class="token operator">+</span> nums<span class="token punctuation">[</span>left<span class="token punctuation">]</span> <span class="token operator">+</span> nums<span class="token punctuation">[</span>right<span class="token punctuation">]</span><span class="token punctuation">;</span></pre></td></tr><tr><td data-num="11"></td><td><pre> <span class="token keyword">if</span> <span class="token punctuation">(</span>sum <span class="token operator">></span> <span class="token number">0</span><span class="token punctuation">)</span> right<span class="token operator">--</span><span class="token punctuation">;</span></pre></td></tr><tr><td data-num="12"></td><td><pre> <span class="token keyword">if</span> <span class="token punctuation">(</span>sum <span class="token operator"><</span> <span class="token number">0</span><span class="token punctuation">)</span> left<span class="token operator">++</span><span class="token punctuation">;</span></pre></td></tr><tr><td data-num="13"></td><td><pre> <span class="token keyword">if</span> <span class="token punctuation">(</span>sum <span class="token operator">==</span> <span class="token number">0</span><span class="token punctuation">)</span> <span class="token punctuation">{</span></pre></td></tr><tr><td data-num="14"></td><td><pre> res<span class="token punctuation">.</span><span class="token function">push_back</span><span class="token punctuation">(</span>vector<span class="token operator"><</span><span class="token keyword">int</span><span class="token operator">></span><span class="token punctuation">{</span>nums<span class="token punctuation">[</span>i<span class="token punctuation">]</span><span class="token punctuation">,</span> nums<span class="token punctuation">[</span>left<span class="token punctuation">]</span><span class="token punctuation">,</span> nums<span class="token punctuation">[</span>right<span class="token punctuation">]</span><span class="token punctuation">}</span><span class="token punctuation">)</span><span class="token punctuation">;</span></pre></td></tr><tr><td data-num="15"></td><td><pre> <span class="token comment">// 将 left 右移（注意，++left 至少执行一次）</span></pre></td></tr><tr><td data-num="16"></td><td><pre> <span class="token keyword">while</span> <span class="token punctuation">(</span>left <span class="token operator"><</span> right <span class="token operator">&&</span> nums<span class="token punctuation">[</span>left<span class="token punctuation">]</span> <span class="token operator">==</span> nums<span class="token punctuation">[</span><span class="token operator">++</span>left<span class="token punctuation">]</span><span class="token punctuation">)</span><span class="token punctuation">;</span></pre></td></tr><tr><td data-num="17"></td><td><pre> <span class="token comment">// 将 right 左移（注意，--right 至少执行一次）</span></pre></td></tr><tr><td data-num="18"></td><td><pre> <span class="token keyword">while</span> <span class="token punctuation">(</span>left <span class="token operator"><</span> right <span class="token operator">&&</span> nums<span class="token punctuation">[</span>right<span class="token punctuation">]</span> <span class="token operator">==</span> nums<span class="token punctuation">[</span><span class="token operator">--</span>right<span class="token punctuation">]</span><span class="token punctuation">)</span><span class="token punctuation">;</span></pre></td></tr><tr><td data-num="19"></td><td><pre> <span class="token punctuation">}</span></pre></td></tr><tr><td data-num="20"></td><td><pre> <span class="token punctuation">}</span></pre></td></tr><tr><td data-num="21"></td><td><pre> <span class="token punctuation">}</span></pre></td></tr><tr><td data-num="22"></td><td><pre> <span class="token keyword">return</span> res<span class="token punctuation">;</span></pre></td></tr><tr><td data-num="23"></td><td><pre><span class="token punctuation">}</span></pre></td></tr></tbody></table></figure><p>需要注意的第一个地方：</p>
<pre><code>if (i > 0 && nums[i] == nums[i - 1]) continue;
</code></pre>
<p>这里不能改成 <code>if (nums[i] == nums[i + 1]) continue;</code> ，否则，以数组 <code>nums = [-1, 0, 1, 2, -1, -4]</code> 为例（排序后， <code>nums = [-4, -1, -1, 0, 1, 2]</code> ），查找时将会遗漏 <code>[-1, -1, 2]</code> 这一个三元组。因为当 <code>i = 1</code> 时 <code>nums[1] == nums[2]</code> 条件成立，因此不会进行双指针查找；而当 <code>i = 2</code> 时，只会查找 <code>i</code> 右侧的元素，故而遗失了对 <code>[-1, -1, x]</code> 这几种情况的查找， <code>x</code> 为第二个 <code>-1</code> 右侧的任意值</p>
<p><!-- 上述的 Method 1 也是类似道理 --></p>
<p>需要注意的第二个地方：</p>
<pre><code>if (sum == 0) {
res.push_back(vector<int>{nums[i], nums[left], nums[right]});
while (left < right && nums[left] == nums[++left]);
while (left < right && nums[right] == nums[--right]);
}
</code></pre>
<p>这里的 <code>++left</code> 和 <code>--right</code> 都至少执行一次：执行一次是因为找到了一个三元组，需要同时移动双指针，以进行下一次的查找；执行多次则是为了跳过重复的 <code>nums[left]</code> 和 <code>nums[right]</code></p>
<p>并且，注意这里的 <code>nums[left] == nums[++left]</code> 语句，将当前 <code>left</code> 对应元素值与 <code>left</code> 右移之后对应元素值进行比较，不能将其改成 <code>nums[++left] == nums[left]</code> ，也不能改成 <code>nums[left] == nums[left++]</code> 。语句 <code>nums[right] == nums[--right]</code> 同理</p>
<p>特别地，第二个地方可以改写成</p>
<pre><code>if (sum == 0) {
res.push_back(vector<int>{nums[i], nums[left], nums[right]});
// 先将 left 移到最靠右的一个 nums[left]
while (left < right && nums[left] == nums[left + 1]) left++;
// 先将 right 移到最靠左的一个 nums[right]
while (left < right && nums[right] == nums[right - 1]) right--;
// 再分别移动一次 left 和 right ，以进行下一次查找
left++;
right--;
}
</code></pre>
<p>另外，当 <code>sum > 0</code> 时，可对 <code>nums[right]</code> 进行去重，以跳过重复的 <code>nums[right]</code> ，即，将 <code>if (sum > 0) right--;</code> 改写为</p>
<pre><code>if (sum > 0)
while (left < right && nums[right] == nums[--right]);
</code></pre>
<p>或者改写为</p>
<pre><code>if (sum > 0) {
right--;
while (left < right && nums[right] == nums[right + 1]) right--;
}
</code></pre>
<p>类似地，当 <code>sum < 0</code> 时，也可对 <code>nums[left]</code> 进行去重，以跳过重复的 <code>nums[left]</code></p>
<p>时间复杂度：<span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>O</mi><mo stretchy="false">(</mo><msup><mi>n</mi><mn>2</mn></msup><mo stretchy="false">)</mo></mrow><annotation encoding="application/x-tex">O(n^2)</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:1.0641em;vertical-align:-0.25em;"></span><span class="mord mathnormal" style="margin-right:0.02778em;">O</span><span class="mopen">(</span><span class="mord"><span class="mord mathnormal">n</span><span class="msupsub"><span class="vlist-t"><span class="vlist-r"><span class="vlist" style="height:0.8141em;"><span style="top:-3.063em;margin-right:0.05em;"><span class="pstrut" style="height:2.7em;"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mtight">2</span></span></span></span></span></span></span></span><span class="mclose">)</span></span></span></span></p>
<ul>
<li>遍历 <code>i</code> ：<span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>O</mi><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow><annotation encoding="application/x-tex">O(n)</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:1em;vertical-align:-0.25em;"></span><span class="mord mathnormal" style="margin-right:0.02778em;">O</span><span class="mopen">(</span><span class="mord mathnormal">n</span><span class="mclose">)</span></span></span></span></li>
<li>双指针查找：<span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>O</mi><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow><annotation encoding="application/x-tex">O(n)</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:1em;vertical-align:-0.25em;"></span><span class="mord mathnormal" style="margin-right:0.02778em;">O</span><span class="mopen">(</span><span class="mord mathnormal">n</span><span class="mclose">)</span></span></span></span></li>
</ul>
<p>空间复杂度：<span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>O</mi><mo stretchy="false">(</mo><mi>log</mi><mo></mo><mi>n</mi><mo stretchy="false">)</mo></mrow><annotation encoding="application/x-tex">O(\log{n})</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:1em;vertical-align:-0.25em;"></span><span class="mord mathnormal" style="margin-right:0.02778em;">O</span><span class="mopen">(</span><span class="mop">lo<span style="margin-right:0.01389em;">g</span></span><span class="mspace" style="margin-right:0.1667em;"></span><span class="mord"><span class="mord mathnormal">n</span></span><span class="mclose">)</span></span></span></span></p>
<ul>
<li>不考虑存储结果的数组</li>
<li>仅考虑排序所需栈空间</li>
</ul>
<p>参考：</p>
<ul>
<li><a target="_blank" rel="noopener" href="https://www.programmercarl.com/0015.%E4%B8%89%E6%95%B0%E4%B9%8B%E5%92%8C.html#%E5%93%88%E5%B8%8C%E8%A7%A3%E6%B3%95">代码随想录：三数之和</a></li>
<li><a target="_blank" rel="noopener" href="https://leetcode-cn.com/problems/3sum/solution/3sumpai-xu-shuang-zhi-zhen-yi-dong-by-jyd/">Krahets：三数之和（排序 + 双指针，易懂图解）</a></li>
</ul>
<h1 id="leetcode-18-四数之和"><a class="anchor" href="#leetcode-18-四数之和">#</a> LeetCode 18. 四数之和</h1>
<p><a target="_blank" rel="noopener" href="https://leetcode-cn.com/problems/4sum/">LeetCode 18. Four Sum</a></p>
<p>给你一个由 <code>n</code> 个整数组成的数组 <code>nums</code> ，和一个目标值 <code>target</code> 。请你找出并返回满足下述全部条件且<strong>不重复</strong>的四元组 <code>[nums[a], nums[b], nums[c], nums[d]]</code> （若两个四元组元素一一对应，则认为两个四元组重复）：</p>
<ul>
<li><code>0 <= a, b, c, d < n</code></li>
<li><code>a</code> 、 <code>b</code> 、 <code>c</code> 和 <code>d</code> <strong>互不相同</strong></li>
<li><code>nums[a] + nums[b] + nums[c] + nums[d] == target</code></li>
</ul>
<p>你可以按 <strong>任意顺序</strong> 返回答案 。</p>
<p></p>
<p><strong>示例 1：</strong></p>
<pre><code>输入：nums = [1,0,-1,0,-2,2], target = 0
输出：[[-2,-1,1,2],[-2,0,0,2],[-1,0,0,1]]
</code></pre>
<p><strong>示例 2：</strong></p>
<pre><code>输入：nums = [2,2,2,2,2], target = 8
输出：[[2,2,2,2]]
</code></pre>
<p></p>
<p><strong>提示：</strong></p>
<ul>
<li><span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mn>1</mn><mo>≤</mo></mrow><annotation encoding="application/x-tex">1 \le</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.7804em;vertical-align:-0.136em;"></span><span class="mord">1</span><span class="mspace" style="margin-right:0.2778em;"></span><span class="mrel">≤</span></span></span></span> <code>nums.length</code> <span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mo>≤</mo><mn>200</mn></mrow><annotation encoding="application/x-tex">\le 200</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.7719em;vertical-align:-0.136em;"></span><span class="mrel">≤</span><span class="mspace" style="margin-right:0.2778em;"></span></span><span class="base"><span class="strut" style="height:0.6444em;"></span><span class="mord">200</span></span></span></span></li>
<li><span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mo>−</mo><mn>1</mn><msup><mn>0</mn><mn>9</mn></msup><mo>≤</mo></mrow><annotation encoding="application/x-tex">- 10^9 \le</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.9501em;vertical-align:-0.136em;"></span><span class="mord">−</span><span class="mord">1</span><span class="mord"><span class="mord">0</span><span class="msupsub"><span class="vlist-t"><span class="vlist-r"><span class="vlist" style="height:0.8141em;"><span style="top:-3.063em;margin-right:0.05em;"><span class="pstrut" style="height:2.7em;"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mtight">9</span></span></span></span></span></span></span></span><span class="mspace" style="margin-right:0.2778em;"></span><span class="mrel">≤</span></span></span></span> <code>nums[i]</code> <span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mo>≤</mo><mn>1</mn><msup><mn>0</mn><mn>9</mn></msup></mrow><annotation encoding="application/x-tex">\le 10^9</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.7719em;vertical-align:-0.136em;"></span><span class="mrel">≤</span><span class="mspace" style="margin-right:0.2778em;"></span></span><span class="base"><span class="strut" style="height:0.8141em;"></span><span class="mord">1</span><span class="mord"><span class="mord">0</span><span class="msupsub"><span class="vlist-t"><span class="vlist-r"><span class="vlist" style="height:0.8141em;"><span style="top:-3.063em;margin-right:0.05em;"><span class="pstrut" style="height:2.7em;"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mtight">9</span></span></span></span></span></span></span></span></span></span></span></li>
<li><span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mo>−</mo><mn>1</mn><msup><mn>0</mn><mn>9</mn></msup><mo>≤</mo></mrow><annotation encoding="application/x-tex">- 10^9 \le</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.9501em;vertical-align:-0.136em;"></span><span class="mord">−</span><span class="mord">1</span><span class="mord"><span class="mord">0</span><span class="msupsub"><span class="vlist-t"><span class="vlist-r"><span class="vlist" style="height:0.8141em;"><span style="top:-3.063em;margin-right:0.05em;"><span class="pstrut" style="height:2.7em;"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mtight">9</span></span></span></span></span></span></span></span><span class="mspace" style="margin-right:0.2778em;"></span><span class="mrel">≤</span></span></span></span> <code>target</code> <span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mo>≤</mo><mn>1</mn><msup><mn>0</mn><mn>9</mn></msup></mrow><annotation encoding="application/x-tex">\le 10^9</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.7719em;vertical-align:-0.136em;"></span><span class="mrel">≤</span><span class="mspace" style="margin-right:0.2778em;"></span></span><span class="base"><span class="strut" style="height:0.8141em;"></span><span class="mord">1</span><span class="mord"><span class="mord">0</span><span class="msupsub"><span class="vlist-t"><span class="vlist-r"><span class="vlist" style="height:0.8141em;"><span style="top:-3.063em;margin-right:0.05em;"><span class="pstrut" style="height:2.7em;"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mtight">9</span></span></span></span></span></span></span></span></span></span></span></li>
</ul>
<h2 id="思路-3"><a class="anchor" href="#思路-3">#</a> 思路</h2>
<p>需注意区分本题与 <strong>LeetCode 454. 四数相加 II</strong></p>
<p>本题的解题方法与 <strong>LeetCode 15. 三数之和</strong> 基本一致，这里采用 排序 + 双指针 解法</p>
<h2 id="method-排序-双指针-2"><a class="anchor" href="#method-排序-双指针-2">#</a> Method: 排序 + 双指针</h2>
<p>解题思路：</p>
<ol>
<li>
<p>对数组 <code>nums</code> 排序</p>
</li>
<li>
<p>采用两个 <code>for</code> 循环分别遍历 <code>i</code> 和 <code>j</code> ，其中， <code>i</code> 从 <code>0</code> 开始遍历， <code>j</code> 从 <code>i + 1</code> 开始遍历（注意，要分别对 <code>nums[i]</code> 和 <code>nums[j]</code> 进行去重）</p>
</li>
<li>
<p>采用双指针法在 <code>[j + 1, nums.size() - 1]</code> 区间内查找所有能与 <code>nums[i]</code> , <code>nums[j]</code> 组成四元组的元素 <code>nums[left]</code> 和 <code>nums[right]</code> ，其中，指针 <code>left</code> 从 <code>j + 1</code> 位置开始向右遍历，指针 <code>right</code> 从 <code>nums.size() - 1</code> 位置开始向左遍历</p>
</li>
</ol>
<p>具体步骤可参考 <strong>LeetCode 15. 三数之和</strong></p>
<p>代码实现：</p>
<figure class="highlight cpp"><figcaption data-lang="C++"></figcaption><table><tbody><tr><td data-num="1"></td><td><pre>vector<span class="token operator"><</span>vector<span class="token operator"><</span><span class="token keyword">int</span><span class="token operator">>></span> <span class="token function">fourSum</span><span class="token punctuation">(</span>vector<span class="token operator"><</span><span class="token keyword">int</span><span class="token operator">></span><span class="token operator">&</span> nums<span class="token punctuation">,</span> <span class="token keyword">int</span> target<span class="token punctuation">)</span> <span class="token punctuation">{</span></pre></td></tr><tr><td data-num="2"></td><td><pre> <span class="token function">sort</span><span class="token punctuation">(</span>nums<span class="token punctuation">.</span><span class="token function">begin</span><span class="token punctuation">(</span><span class="token punctuation">)</span><span class="token punctuation">,</span> nums<span class="token punctuation">.</span><span class="token function">end</span><span class="token punctuation">(</span><span class="token punctuation">)</span><span class="token punctuation">)</span><span class="token punctuation">;</span></pre></td></tr><tr><td data-num="3"></td><td><pre> vector<span class="token operator"><</span>vector<span class="token operator"><</span><span class="token keyword">int</span><span class="token operator">>></span> res<span class="token punctuation">;</span></pre></td></tr><tr><td data-num="4"></td><td><pre> <span class="token keyword">for</span> <span class="token punctuation">(</span><span class="token keyword">int</span> i <span class="token operator">=</span> <span class="token number">0</span><span class="token punctuation">;</span> i <span class="token operator"><</span> nums<span class="token punctuation">.</span><span class="token function">size</span><span class="token punctuation">(</span><span class="token punctuation">)</span><span class="token punctuation">;</span> i<span class="token operator">++</span><span class="token punctuation">)</span> <span class="token punctuation">{</span></pre></td></tr><tr><td data-num="5"></td><td><pre> <span class="token keyword">if</span> <span class="token punctuation">(</span>i <span class="token operator">></span> <span class="token number">0</span> <span class="token operator">&&</span> nums<span class="token punctuation">[</span>i<span class="token punctuation">]</span> <span class="token operator">==</span> nums<span class="token punctuation">[</span>i <span class="token operator">-</span> <span class="token number">1</span><span class="token punctuation">]</span><span class="token punctuation">)</span> <span class="token keyword">continue</span><span class="token punctuation">;</span> <span class="token comment">// 对 nums [i] 去重</span></pre></td></tr><tr><td data-num="6"></td><td><pre> <span class="token keyword">for</span> <span class="token punctuation">(</span><span class="token keyword">int</span> j <span class="token operator">=</span> i <span class="token operator">+</span> <span class="token number">1</span><span class="token punctuation">;</span> j <span class="token operator"><</span> nums<span class="token punctuation">.</span><span class="token function">size</span><span class="token punctuation">(</span><span class="token punctuation">)</span><span class="token punctuation">;</span> j<span class="token operator">++</span><span class="token punctuation">)</span> <span class="token punctuation">{</span></pre></td></tr><tr><td data-num="7"></td><td><pre> <span class="token keyword">if</span> <span class="token punctuation">(</span>j <span class="token operator">></span> i <span class="token operator">+</span> <span class="token number">1</span> <span class="token operator">&&</span> nums<span class="token punctuation">[</span>j<span class="token punctuation">]</span> <span class="token operator">==</span> nums<span class="token punctuation">[</span>j <span class="token operator">-</span> <span class="token number">1</span><span class="token punctuation">]</span><span class="token punctuation">)</span> <span class="token keyword">continue</span><span class="token punctuation">;</span> <span class="token comment">// 对 nums [j] 去重</span></pre></td></tr><tr><td data-num="8"></td><td><pre> <span class="token keyword">int</span> left <span class="token operator">=</span> j <span class="token operator">+</span> <span class="token number">1</span><span class="token punctuation">,</span> right <span class="token operator">=</span> nums<span class="token punctuation">.</span><span class="token function">size</span><span class="token punctuation">(</span><span class="token punctuation">)</span> <span class="token operator">-</span> <span class="token number">1</span><span class="token punctuation">;</span></pre></td></tr><tr><td data-num="9"></td><td><pre> <span class="token keyword">int</span> temp <span class="token operator">=</span> nums<span class="token punctuation">[</span>i<span class="token punctuation">]</span> <span class="token operator">+</span> nums<span class="token punctuation">[</span>j<span class="token punctuation">]</span><span class="token punctuation">;</span> <span class="token comment">// 直接计算 nums [i] + nums [j] + nums [left] + nums [right] 会溢出</span></pre></td></tr><tr><td data-num="10"></td><td><pre> <span class="token keyword">while</span> <span class="token punctuation">(</span>left <span class="token operator"><</span> right<span class="token punctuation">)</span> <span class="token punctuation">{</span></pre></td></tr><tr><td data-num="11"></td><td><pre> <span class="token keyword">if</span> <span class="token punctuation">(</span>nums<span class="token punctuation">[</span>left<span class="token punctuation">]</span> <span class="token operator">+</span> nums<span class="token punctuation">[</span>right<span class="token punctuation">]</span> <span class="token operator"><</span> target <span class="token operator">-</span> temp<span class="token punctuation">)</span> left<span class="token operator">++</span><span class="token punctuation">;</span></pre></td></tr><tr><td data-num="12"></td><td><pre> <span class="token keyword">else</span> <span class="token keyword">if</span> <span class="token punctuation">(</span>nums<span class="token punctuation">[</span>left<span class="token punctuation">]</span> <span class="token operator">+</span> nums<span class="token punctuation">[</span>right<span class="token punctuation">]</span> <span class="token operator">></span> target <span class="token operator">-</span> temp<span class="token punctuation">)</span> right<span class="token operator">--</span><span class="token punctuation">;</span></pre></td></tr><tr><td data-num="13"></td><td><pre> <span class="token keyword">else</span> <span class="token punctuation">{</span></pre></td></tr><tr><td data-num="14"></td><td><pre> res<span class="token punctuation">.</span><span class="token function">push_back</span><span class="token punctuation">(</span><span class="token punctuation">{</span>nums<span class="token punctuation">[</span>i<span class="token punctuation">]</span><span class="token punctuation">,</span> nums<span class="token punctuation">[</span>j<span class="token punctuation">]</span><span class="token punctuation">,</span> nums<span class="token punctuation">[</span>left<span class="token punctuation">]</span><span class="token punctuation">,</span> nums<span class="token punctuation">[</span>right<span class="token punctuation">]</span><span class="token punctuation">}</span><span class="token punctuation">)</span><span class="token punctuation">;</span></pre></td></tr><tr><td data-num="15"></td><td><pre> <span class="token keyword">while</span> <span class="token punctuation">(</span>left <span class="token operator"><</span> right <span class="token operator">&&</span> nums<span class="token punctuation">[</span>left<span class="token punctuation">]</span> <span class="token operator">==</span> nums<span class="token punctuation">[</span><span class="token operator">++</span>left<span class="token punctuation">]</span><span class="token punctuation">)</span><span class="token punctuation">;</span> <span class="token comment">// 对 nums [left] 去重</span></pre></td></tr><tr><td data-num="16"></td><td><pre> <span class="token keyword">while</span> <span class="token punctuation">(</span>left <span class="token operator"><</span> right <span class="token operator">&&</span> nums<span class="token punctuation">[</span>right<span class="token punctuation">]</span> <span class="token operator">==</span> nums<span class="token punctuation">[</span><span class="token operator">--</span>right<span class="token punctuation">]</span><span class="token punctuation">)</span><span class="token punctuation">;</span> <span class="token comment">// 对 nums [right] 去重</span></pre></td></tr><tr><td data-num="17"></td><td><pre> <span class="token punctuation">}</span></pre></td></tr><tr><td data-num="18"></td><td><pre> <span class="token punctuation">}</span></pre></td></tr><tr><td data-num="19"></td><td><pre> <span class="token punctuation">}</span></pre></td></tr><tr><td data-num="20"></td><td><pre> <span class="token punctuation">}</span></pre></td></tr><tr><td data-num="21"></td><td><pre> <span class="token keyword">return</span> res<span class="token punctuation">;</span></pre></td></tr><tr><td data-num="22"></td><td><pre><span class="token punctuation">}</span></pre></td></tr></tbody></table></figure><p>其中，需要注意 <code>nums[i] + nums[j] + nums[left] + nums[right]</code> 会溢出，因此，不能直接将 <code>nums[i] + nums[j] + nums[left] + nums[right]</code> 与 <code>target</code> 比较，可比较 <code>nums[left] + nums[right]</code> 与 <code>target - nums[i] - nums[j]</code></p>
<p>另外，当 <code>temp2 < target - temp1</code> 时，可跳过重复的 <code>nums[left]</code> ，当 <code>temp2 > target - temp1</code> 时，可跳过重复的 <code>nums[right]</code> ，即，将第 12 至 13 行代码改写为：</p>
<pre><code>if (temp2 < target - temp1)
while (left < right && nums[left] == nums[++left]);
else if (temp2 > target - temp1)
while (left < right && nums[right] == nums[--right]);
</code></pre>
<p>具体可见 <strong>LeetCode 15. 三数之和</strong></p>
<p>时间复杂度：<span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>O</mi><mo stretchy="false">(</mo><msup><mi>n</mi><mn>3</mn></msup><mo stretchy="false">)</mo></mrow><annotation encoding="application/x-tex">O(n^3)</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:1.0641em;vertical-align:-0.25em;"></span><span class="mord mathnormal" style="margin-right:0.02778em;">O</span><span class="mopen">(</span><span class="mord"><span class="mord mathnormal">n</span><span class="msupsub"><span class="vlist-t"><span class="vlist-r"><span class="vlist" style="height:0.8141em;"><span style="top:-3.063em;margin-right:0.05em;"><span class="pstrut" style="height:2.7em;"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mtight">3</span></span></span></span></span></span></span></span><span class="mclose">)</span></span></span></span></p>
<ul>
<li>遍历 <code>i</code> 和 <code>j</code> ：<span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>O</mi><mo stretchy="false">(</mo><msup><mi>n</mi><mn>2</mn></msup><mo stretchy="false">)</mo></mrow><annotation encoding="application/x-tex">O(n^2)</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:1.0641em;vertical-align:-0.25em;"></span><span class="mord mathnormal" style="margin-right:0.02778em;">O</span><span class="mopen">(</span><span class="mord"><span class="mord mathnormal">n</span><span class="msupsub"><span class="vlist-t"><span class="vlist-r"><span class="vlist" style="height:0.8141em;"><span style="top:-3.063em;margin-right:0.05em;"><span class="pstrut" style="height:2.7em;"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mtight">2</span></span></span></span></span></span></span></span><span class="mclose">)</span></span></span></span></li>
<li>双指针查找：<span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>O</mi><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow><annotation encoding="application/x-tex">O(n)</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:1em;vertical-align:-0.25em;"></span><span class="mord mathnormal" style="margin-right:0.02778em;">O</span><span class="mopen">(</span><span class="mord mathnormal">n</span><span class="mclose">)</span></span></span></span></li>
</ul>
<p>空间复杂度：<span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>O</mi><mo stretchy="false">(</mo><mi>log</mi><mo></mo><mi>n</mi><mo stretchy="false">)</mo></mrow><annotation encoding="application/x-tex">O(\log{n})</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:1em;vertical-align:-0.25em;"></span><span class="mord mathnormal" style="margin-right:0.02778em;">O</span><span class="mopen">(</span><span class="mop">lo<span style="margin-right:0.01389em;">g</span></span><span class="mspace" style="margin-right:0.1667em;"></span><span class="mord"><span class="mord mathnormal">n</span></span><span class="mclose">)</span></span></span></span> ，这里不考虑存储结果的数组，仅考虑排序所需栈空间</p>
<p>参考：<a target="_blank" rel="noopener" href="https://www.programmercarl.com/0018.%E5%9B%9B%E6%95%B0%E4%B9%8B%E5%92%8C.html">代码随想录：四数之和</a></p>
<h1 id="leetcode-202-快乐数"><a class="anchor" href="#leetcode-202-快乐数">#</a> LeetCode 202. 快乐数</h1>
<p><a target="_blank" rel="noopener" href="https://leetcode-cn.com/problems/happy-number/">LeetCode 202. Happy Number</a></p>
<p>编写一个算法来判断一个数 <code>n</code> 是不是快乐数。</p>
<p><strong>快乐数</strong> 定义为：</p>
<ul>
<li>对于一个正整数，每一次将该数替换为它每个位置上的数字的平方和。</li>
<li>然后重复这个过程直到这个数变为 1，也可能是 <strong>无限循环</strong> 但始终变不到 1。</li>
<li>如果这个过程 <strong>结果为 1</strong>，那么这个数就是快乐数。</li>
</ul>
<p>如果 <code>n</code> 是 快乐数 就返回 <code>true</code> ；不是，则返回 <code>false</code> 。</p>
<p></p>
<p><strong>示例 1：</strong></p>
<pre><code>输入：n = 19
输出：true
解释：
1 + 81 = 82
64 + 4 = 68
36 + 64 = 100
1 + 0 + 0 = 1
</code></pre>
<p><strong>示例 2：</strong></p>
<pre><code>输入：n = 2
输出：false
</code></pre>
<p></p>
<p><strong>提示：</strong></p>
<ul>
<li><span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mn>1</mn><mo>≤</mo></mrow><annotation encoding="application/x-tex">1 \le</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.7804em;vertical-align:-0.136em;"></span><span class="mord">1</span><span class="mspace" style="margin-right:0.2778em;"></span><span class="mrel">≤</span></span></span></span> <code>n</code> <span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mo>≤</mo><msup><mn>2</mn><mn>31</mn></msup><mo>−</mo><mn>1</mn></mrow><annotation encoding="application/x-tex">\le 2^{31} - 1</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.7719em;vertical-align:-0.136em;"></span><span class="mrel">≤</span><span class="mspace" style="margin-right:0.2778em;"></span></span><span class="base"><span class="strut" style="height:0.8974em;vertical-align:-0.0833em;"></span><span class="mord"><span class="mord">2</span><span class="msupsub"><span class="vlist-t"><span class="vlist-r"><span class="vlist" style="height:0.8141em;"><span style="top:-3.063em;margin-right:0.05em;"><span class="pstrut" style="height:2.7em;"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mtight"><span class="mord mtight">31</span></span></span></span></span></span></span></span></span><span class="mspace" style="margin-right:0.2222em;"></span><span class="mbin">−</span><span class="mspace" style="margin-right:0.2222em;"></span></span><span class="base"><span class="strut" style="height:0.6444em;"></span><span class="mord">1</span></span></span></span></li>
</ul>
<h2 id="思路-4"><a class="anchor" href="#思路-4">#</a> 思路</h2>
<p>注意到题目说明：</p>
<ul>
<li>若各位平方和最终等于 1 ，则 <code>n</code> 为快乐数</li>
<li>若各位平方和的值循环出现，则 <code>n</code> 不是快乐数</li>
</ul>
<p>因此，本题的解题关键在于，在求和过程中，判断数字是否重复出现</p>
<ul>
<li>若重复出现，则该数字必定不是快乐数</li>
<li>若求和结果为 1，则该数字是快乐数</li>
</ul>
<h2 id="method-1-哈希-set"><a class="anchor" href="#method-1-哈希-set">#</a> Method 1: 哈希 set</h2>
<p><a target="_blank" rel="noopener" href="http://www.cplusplus.com/reference/unordered_set/unordered_set/?kw=unordered_set">std::unordered_set</a></p>
<p>考虑用哈希表 <code>unordered_set</code> 记录求和过程中出现的数字，并判断其是否重复出现</p>
<p>另，需注意 <code>n</code> 的各位数字平方和的计算</p>
<ul>
<li>计算个位数字的平方</li>
<li>取个位以外的部分</li>
<li>重复以上两步</li>
</ul>
<p>代码实现：</p>
<figure class="highlight cpp"><figcaption data-lang="C++"></figcaption><table><tbody><tr><td data-num="1"></td><td><pre><span class="token keyword">int</span> <span class="token function">getSum</span><span class="token punctuation">(</span><span class="token keyword">int</span> n<span class="token punctuation">)</span> <span class="token punctuation">{</span> <span class="token comment">// 计算 n 中各位的平方和</span></pre></td></tr><tr><td data-num="2"></td><td><pre> <span class="token keyword">int</span> sum <span class="token operator">=</span> <span class="token number">0</span><span class="token punctuation">;</span></pre></td></tr><tr><td data-num="3"></td><td><pre> <span class="token keyword">while</span> <span class="token punctuation">(</span>n<span class="token punctuation">)</span> <span class="token punctuation">{</span></pre></td></tr><tr><td data-num="4"></td><td><pre> sum <span class="token operator">+=</span> <span class="token punctuation">(</span>n <span class="token operator">%</span> <span class="token number">10</span><span class="token punctuation">)</span> <span class="token operator">*</span> <span class="token punctuation">(</span>n <span class="token operator">%</span> <span class="token number">10</span><span class="token punctuation">)</span><span class="token punctuation">;</span> <span class="token comment">// 取 n 的个位数，求其平方，再累加到 sum 上</span></pre></td></tr><tr><td data-num="5"></td><td><pre> n <span class="token operator">/=</span> <span class="token number">10</span><span class="token punctuation">;</span> <span class="token comment">// 取个位数以外的部分</span></pre></td></tr><tr><td data-num="6"></td><td><pre> <span class="token punctuation">}</span></pre></td></tr><tr><td data-num="7"></td><td><pre> <span class="token keyword">return</span> sum<span class="token punctuation">;</span></pre></td></tr><tr><td data-num="8"></td><td><pre><span class="token punctuation">}</span></pre></td></tr><tr><td data-num="9"></td><td><pre></pre></td></tr><tr><td data-num="10"></td><td><pre><span class="token keyword">bool</span> <span class="token function">isHappy</span><span class="token punctuation">(</span><span class="token keyword">int</span> n<span class="token punctuation">)</span> <span class="token punctuation">{</span></pre></td></tr><tr><td data-num="11"></td><td><pre> unordered_set<span class="token operator"><</span><span class="token keyword">int</span><span class="token operator">></span> record<span class="token punctuation">;</span></pre></td></tr><tr><td data-num="12"></td><td><pre> <span class="token keyword">while</span> <span class="token punctuation">(</span><span class="token number">1</span><span class="token punctuation">)</span> <span class="token punctuation">{</span></pre></td></tr><tr><td data-num="13"></td><td><pre> <span class="token keyword">int</span> sum <span class="token operator">=</span> <span class="token function">getSum</span><span class="token punctuation">(</span>n<span class="token punctuation">)</span><span class="token punctuation">;</span></pre></td></tr><tr><td data-num="14"></td><td><pre> <span class="token keyword">if</span> <span class="token punctuation">(</span>sum <span class="token operator">==</span> <span class="token number">1</span><span class="token punctuation">)</span></pre></td></tr><tr><td data-num="15"></td><td><pre> <span class="token keyword">return</span> <span class="token number">1</span><span class="token punctuation">;</span></pre></td></tr><tr><td data-num="16"></td><td><pre> <span class="token keyword">if</span> <span class="token punctuation">(</span>record<span class="token punctuation">.</span><span class="token function">find</span><span class="token punctuation">(</span>sum<span class="token punctuation">)</span> <span class="token operator">!=</span> record<span class="token punctuation">.</span><span class="token function">end</span><span class="token punctuation">(</span><span class="token punctuation">)</span><span class="token punctuation">)</span> <span class="token comment">//sum 重复出现</span></pre></td></tr><tr><td data-num="17"></td><td><pre> <span class="token keyword">return</span> <span class="token number">0</span><span class="token punctuation">;</span></pre></td></tr><tr><td data-num="18"></td><td><pre> <span class="token keyword">else</span></pre></td></tr><tr><td data-num="19"></td><td><pre> record<span class="token punctuation">.</span><span class="token function">insert</span><span class="token punctuation">(</span>sum<span class="token punctuation">)</span><span class="token punctuation">;</span> <span class="token comment">//sum 第一次出现</span></pre></td></tr><tr><td data-num="20"></td><td><pre> n <span class="token operator">=</span> sum<span class="token punctuation">;</span> <span class="token comment">// 更新当前计算数字</span></pre></td></tr><tr><td data-num="21"></td><td><pre> <span class="token punctuation">}</span></pre></td></tr><tr><td data-num="22"></td><td><pre><span class="token punctuation">}</span></pre></td></tr></tbody></table></figure><p><a target="_blank" rel="noopener" href="https://www.programmercarl.com/0202.%E5%BF%AB%E4%B9%90%E6%95%B0.html">代码随想录：快乐数</a></p>
<p>时间复杂度：<span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>O</mi><mo stretchy="false">(</mo><mi>log</mi><mo></mo><mi>n</mi><mo stretchy="false">)</mo></mrow><annotation encoding="application/x-tex">O(\log{n})</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:1em;vertical-align:-0.25em;"></span><span class="mord mathnormal" style="margin-right:0.02778em;">O</span><span class="mopen">(</span><span class="mop">lo<span style="margin-right:0.01389em;">g</span></span><span class="mspace" style="margin-right:0.1667em;"></span><span class="mord"><span class="mord mathnormal">n</span></span><span class="mclose">)</span></span></span></span></p>
<ul>
<li>查找给定数字的下一个值的时间复杂度为 <span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>O</mi><mo stretchy="false">(</mo><mi>log</mi><mo></mo><mi>n</mi><mo stretchy="false">)</mo></mrow><annotation encoding="application/x-tex">O(\log{n})</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:1em;vertical-align:-0.25em;"></span><span class="mord mathnormal" style="margin-right:0.02778em;">O</span><span class="mopen">(</span><span class="mop">lo<span style="margin-right:0.01389em;">g</span></span><span class="mspace" style="margin-right:0.1667em;"></span><span class="mord"><span class="mord mathnormal">n</span></span><span class="mclose">)</span></span></span></span> ，因为数字的位数由 <span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>O</mi><mo stretchy="false">(</mo><msub><mrow><mi>log</mi><mo></mo></mrow><mn>10</mn></msub><mi>n</mi><mo stretchy="false">)</mo><mo>=</mo><mi>O</mi><mo stretchy="false">(</mo><mi>log</mi><mo></mo><mi>n</mi><mo stretchy="false">)</mo></mrow><annotation encoding="application/x-tex">O(\log_{10}{n}) = O(\log{n})</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:1em;vertical-align:-0.25em;"></span><span class="mord mathnormal" style="margin-right:0.02778em;">O</span><span class="mopen">(</span><span class="mop"><span class="mop">lo<span style="margin-right:0.01389em;">g</span></span><span class="msupsub"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist" style="height:0.207em;"><span style="top:-2.4559em;margin-right:0.05em;"><span class="pstrut" style="height:2.7em;"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mtight"><span class="mord mtight">10</span></span></span></span></span><span class="vlist-s"></span></span><span class="vlist-r"><span class="vlist" style="height:0.2441em;"><span></span></span></span></span></span></span><span class="mspace" style="margin-right:0.1667em;"></span><span class="mord"><span class="mord mathnormal">n</span></span><span class="mclose">)</span><span class="mspace" style="margin-right:0.2778em;"></span><span class="mrel">=</span><span class="mspace" style="margin-right:0.2778em;"></span></span><span class="base"><span class="strut" style="height:1em;vertical-align:-0.25em;"></span><span class="mord mathnormal" style="margin-right:0.02778em;">O</span><span class="mopen">(</span><span class="mop">lo<span style="margin-right:0.01389em;">g</span></span><span class="mspace" style="margin-right:0.1667em;"></span><span class="mord"><span class="mord mathnormal">n</span></span><span class="mclose">)</span></span></span></span> 确定</li>
<li>总体的时间复杂度还需考虑循环过程中的数字个数，这里简单起见仅考虑计算下一个值的时间复杂度</li>
</ul>
<p>空间复杂度：<span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>O</mi><mo stretchy="false">(</mo><mi>log</mi><mo></mo><mi>n</mi><mo stretchy="false">)</mo></mrow><annotation encoding="application/x-tex">O(\log{n})</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:1em;vertical-align:-0.25em;"></span><span class="mord mathnormal" style="margin-right:0.02778em;">O</span><span class="mopen">(</span><span class="mop">lo<span style="margin-right:0.01389em;">g</span></span><span class="mspace" style="margin-right:0.1667em;"></span><span class="mord"><span class="mord mathnormal">n</span></span><span class="mclose">)</span></span></span></span></p>
<h2 id="method-2-快慢指针"><a class="anchor" href="#method-2-快慢指针">#</a> Method 2: 快慢指针</h2>
<p>可以采用类似于 <strong>LeetCode 142. 环形链表 II</strong> 的方法，定义快慢指针，判断是否存在环</p>
<ul>
<li>若存在环，则快慢指针一定会在环内相遇，即，数 <code>n</code> 不是快乐数</li>
<li>否则，快指针会比慢指针先到达数字 1 ，即，数 <code>n</code> 是快乐数</li>
</ul>
<p><a target="_blank" rel="noopener" href="https://leetcode-cn.com/problems/happy-number/solution/kuai-le-shu-by-leetcode-solution/">力扣官方题解：快乐数</a></p>
<h1 id="leetcode-242-有效的字母异位词"><a class="anchor" href="#leetcode-242-有效的字母异位词">#</a> LeetCode 242. 有效的字母异位词</h1>
<p><a target="_blank" rel="noopener" href="https://leetcode-cn.com/problems/valid-anagram/">LeetCode 242. Valid Anagram</a></p>
<p>给定两个字符串 <code>s</code> 和 <code>t</code> ，编写一个函数来判断 <code>t</code> 是否是 <code>s</code> 的字母异位词。</p>
<p>注意：若 <code>s</code> 和 <code>t</code> 中每个字符出现的次数都相同，则称 <code>s</code> 和 <code>t</code> 互为字母异位词。</p>
<p></p>
<p><strong>示例 1：</strong></p>
<pre><code>输入：s = "anagram", t = "nagaram"
输出：true
</code></pre>
<p><strong>示例 2：</strong></p>
<pre><code>输入：s = "rat", t = "car"
输出：false
</code></pre>
<p></p>
<p><strong>提示：</strong></p>
<ul>
<li><span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mn>1</mn><mo>≤</mo></mrow><annotation encoding="application/x-tex">1 \le</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.7804em;vertical-align:-0.136em;"></span><span class="mord">1</span><span class="mspace" style="margin-right:0.2778em;"></span><span class="mrel">≤</span></span></span></span> <code>s.length</code> , <code>t.length</code> <span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mo>≤</mo><mn>5</mn><mo>×</mo><mn>1</mn><msup><mn>0</mn><mn>4</mn></msup></mrow><annotation encoding="application/x-tex">\le 5 \times 10^4</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.7719em;vertical-align:-0.136em;"></span><span class="mrel">≤</span><span class="mspace" style="margin-right:0.2778em;"></span></span><span class="base"><span class="strut" style="height:0.7278em;vertical-align:-0.0833em;"></span><span class="mord">5</span><span class="mspace" style="margin-right:0.2222em;"></span><span class="mbin">×</span><span class="mspace" style="margin-right:0.2222em;"></span></span><span class="base"><span class="strut" style="height:0.8141em;"></span><span class="mord">1</span><span class="mord"><span class="mord">0</span><span class="msupsub"><span class="vlist-t"><span class="vlist-r"><span class="vlist" style="height:0.8141em;"><span style="top:-3.063em;margin-right:0.05em;"><span class="pstrut" style="height:2.7em;"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mtight">4</span></span></span></span></span></span></span></span></span></span></span></li>
<li><code>s</code> 和 <code>t</code> 仅包含小写字母</li>
</ul>
<p><strong>进阶</strong>：如果输入字符串包含 unicode 字符怎么办？你能否调整你的解法来应对这种情况？</p>
<h2 id="思路-5"><a class="anchor" href="#思路-5">#</a> 思路</h2>
<p>字母异位词，等价于，两个字符串中字符出现的种类和次数均相等</p>
<p>利用数组 <code>cnt</code> 来实现一个简单的哈希表，将字符 a 到 z 映射为数组的下标 0 到 25 ，数组的元素值表示相应字符在字符串 <code>s</code> 和 <code>t</code> 里出现的次数的差值</p>
<p>若 <code>cnt</code> 所有元素值均为 0，字符串 <code>s</code> 和 <code>t</code> 是字母异位词</p>
<p>代码实现：</p>
<figure class="highlight cpp"><figcaption data-lang="C++"></figcaption><table><tbody><tr><td data-num="1"></td><td><pre><span class="token keyword">bool</span> <span class="token function">isAnagram</span><span class="token punctuation">(</span>string s<span class="token punctuation">,</span> string t<span class="token punctuation">)</span> <span class="token punctuation">{</span></pre></td></tr><tr><td data-num="2"></td><td><pre> vector<span class="token operator"><</span><span class="token keyword">int</span><span class="token operator">></span> <span class="token function">cnt</span><span class="token punctuation">(</span><span class="token number">26</span><span class="token punctuation">,</span> <span class="token number">0</span><span class="token punctuation">)</span><span class="token punctuation">;</span> <span class="token comment">// 计算 s 和 t 各个字母出现次数的差值</span></pre></td></tr><tr><td data-num="3"></td><td><pre> <span class="token keyword">for</span> <span class="token punctuation">(</span><span class="token keyword">auto</span> c <span class="token operator">:</span> s<span class="token punctuation">)</span></pre></td></tr><tr><td data-num="4"></td><td><pre> cnt<span class="token punctuation">[</span>c <span class="token operator">-</span> <span class="token char">'a'</span><span class="token punctuation">]</span><span class="token operator">++</span><span class="token punctuation">;</span> <span class="token comment">// 记录字符串 s 中字符出现的次数</span></pre></td></tr><tr><td data-num="5"></td><td><pre> <span class="token keyword">for</span> <span class="token punctuation">(</span><span class="token keyword">auto</span> c <span class="token operator">:</span> t<span class="token punctuation">)</span></pre></td></tr><tr><td data-num="6"></td><td><pre> cnt<span class="token punctuation">[</span>c <span class="token operator">-</span> <span class="token char">'a'</span><span class="token punctuation">]</span><span class="token operator">--</span><span class="token punctuation">;</span> <span class="token comment">// 记录字符串 t 中字符出现次数的相反数</span></pre></td></tr><tr><td data-num="7"></td><td><pre> <span class="token keyword">for</span> <span class="token punctuation">(</span><span class="token keyword">int</span> i <span class="token operator">=</span> <span class="token number">0</span><span class="token punctuation">;</span> i <span class="token operator"><</span> <span class="token number">26</span><span class="token punctuation">;</span> i<span class="token operator">++</span><span class="token punctuation">)</span></pre></td></tr><tr><td data-num="8"></td><td><pre> <span class="token keyword">if</span> <span class="token punctuation">(</span>cnt<span class="token punctuation">[</span>i<span class="token punctuation">]</span> <span class="token operator">!=</span> <span class="token number">0</span><span class="token punctuation">)</span> <span class="token comment">//cnt 元素不为 0 ，则字符串 s 和 t 中的字符不同，不是字母异位词</span></pre></td></tr><tr><td data-num="9"></td><td><pre> <span class="token keyword">return</span> <span class="token number">0</span><span class="token punctuation">;</span></pre></td></tr><tr><td data-num="10"></td><td><pre> <span class="token keyword">return</span> <span class="token number">1</span><span class="token punctuation">;</span> <span class="token comment">// 所有字符出现次数相同，是 s 和 t 是字母异位词</span></pre></td></tr><tr><td data-num="11"></td><td><pre><span class="token punctuation">}</span></pre></td></tr></tbody></table></figure><p>时间复杂度：<span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>O</mi><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow><annotation encoding="application/x-tex">O(n)</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:1em;vertical-align:-0.25em;"></span><span class="mord mathnormal" style="margin-right:0.02778em;">O</span><span class="mopen">(</span><span class="mord mathnormal">n</span><span class="mclose">)</span></span></span></span></p>
<p>空间复杂度：<span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>O</mi><mo stretchy="false">(</mo><mn>1</mn><mo stretchy="false">)</mo></mrow><annotation encoding="application/x-tex">O(1)</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:1em;vertical-align:-0.25em;"></span><span class="mord mathnormal" style="margin-right:0.02778em;">O</span><span class="mopen">(</span><span class="mord">1</span><span class="mclose">)</span></span></span></span>，常量大小的辅助数组</p>
<blockquote>
<p>这一题使用数组来实现哈希法，是因为题目限制了只有 26 个小写字母，哈希值比较集中</p>
</blockquote>
<h1 id="leetcode-349-两个数组的交集"><a class="anchor" href="#leetcode-349-两个数组的交集">#</a> LeetCode 349. 两个数组的交集</h1>
<p><a target="_blank" rel="noopener" href="https://leetcode-cn.com/problems/intersection-of-two-arrays/">LeetCode 349. Intersection of Two Arrays</a></p>
<p>给定两个数组 <code>nums1</code> 和 <code>nums2</code> ，返回 它们的交集 。输出结果中的每个元素一定是 唯一 的。我们可以 <strong>不考虑输出结果的顺序</strong> 。</p>
<p><strong>示例 1：</strong></p>
<pre><code>输入：nums1 = [1,2,2,1], nums2 = [2,2]
输出：[2]
</code></pre>
<p><strong>示例 2：</strong></p>
<pre><code>输入：nums1 = [4,9,5], nums2 = [9,4,9,8,4]
输出：[9,4]
解释：[4,9] 也是可通过的
</code></pre>
<p></p>
<p><strong>提示：</strong></p>
<ul>
<li><span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mn>1</mn><mo>≤</mo></mrow><annotation encoding="application/x-tex">1 \le</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.7804em;vertical-align:-0.136em;"></span><span class="mord">1</span><span class="mspace" style="margin-right:0.2778em;"></span><span class="mrel">≤</span></span></span></span> <code>nums1.length</code> , <code>nums2.length</code> <span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mo>≤</mo><mn>1000</mn></mrow><annotation encoding="application/x-tex">\le 1000</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.7719em;vertical-align:-0.136em;"></span><span class="mrel">≤</span><span class="mspace" style="margin-right:0.2778em;"></span></span><span class="base"><span class="strut" style="height:0.6444em;"></span><span class="mord">1000</span></span></span></span></li>
<li><span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mn>0</mn><mo>≤</mo></mrow><annotation encoding="application/x-tex">0 \le</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.7804em;vertical-align:-0.136em;"></span><span class="mord">0</span><span class="mspace" style="margin-right:0.2778em;"></span><span class="mrel">≤</span></span></span></span> <code>nums1[i]</code> , <code>nums2[i]</code> <span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mo>≤</mo><mn>1000</mn></mrow><annotation encoding="application/x-tex">\le 1000</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.7719em;vertical-align:-0.136em;"></span><span class="mrel">≤</span><span class="mspace" style="margin-right:0.2778em;"></span></span><span class="base"><span class="strut" style="height:0.6444em;"></span><span class="mord">1000</span></span></span></span></li>
</ul>
<h2 id="思路-6"><a class="anchor" href="#思路-6">#</a> 思路</h2>
<p>由于数组 <code>nums1</code> 和 <code>nums2</code> 的元素值可能比较分散，使用数组来实现哈希法会造成空间的浪费，故而可以采用 <code>set</code></p>
<p>注意到题目要求输出的每个元素必须唯一，且不考虑输出结果的顺序，故而选用 <code>unordered_set</code></p>
<blockquote>
<p>注：相比于数组而言，直接使用 set 不仅占用空间比数组大，而且速度慢。因此，在 数组元素大小有限 且 哈希值比较集中 时，应尽量用 数组 ，而如果哈希值比较少、特别分散、跨度非常大，则考虑用 set</p>
</blockquote>
<h2 id="unordered_set"><a class="anchor" href="#unordered_set">#</a> unordered_set</h2>
<p>解题步骤：</p>
<ol>
<li>
<p>定义 <code>unordered_set</code> 容器 <code>record</code> 和 <code>ans</code> ，前者存放第一个数组 <code>nums1</code> 的元素，后者记录两数组的交集</p>
</li>
<li>
<p>遍历数组 <code>nums2</code> ：若 <code>nums2</code> 数组的元素 <code>a</code> 能够在 <code>record</code> 中找到（即， <code>record.find(a) != record.end()</code> ），则说明 <code>a</code> 是两数组的公共元素，将其添加到 <code>ans</code> 中</p>
</li>
</ol>
<p>代码实现：</p>
<figure class="highlight cpp"><figcaption data-lang="C++"></figcaption><table><tbody><tr><td data-num="1"></td><td><pre>vector<span class="token operator"><</span><span class="token keyword">int</span><span class="token operator">></span> <span class="token function">intersection</span><span class="token punctuation">(</span>vector<span class="token operator"><</span><span class="token keyword">int</span><span class="token operator">></span><span class="token operator">&</span> nums1<span class="token punctuation">,</span> vector<span class="token operator"><</span><span class="token keyword">int</span><span class="token operator">></span><span class="token operator">&</span> nums2<span class="token punctuation">)</span> <span class="token punctuation">{</span></pre></td></tr><tr><td data-num="2"></td><td><pre> unordered_set<span class="token operator"><</span><span class="token keyword">int</span><span class="token operator">></span> <span class="token function">record</span><span class="token punctuation">(</span>nums1<span class="token punctuation">.</span><span class="token function">begin</span><span class="token punctuation">(</span><span class="token punctuation">)</span><span class="token punctuation">,</span> nums1<span class="token punctuation">.</span><span class="token function">end</span><span class="token punctuation">(</span><span class="token punctuation">)</span><span class="token punctuation">)</span><span class="token punctuation">;</span> <span class="token comment">// 将 nums1 拷贝到 unordered_set 容器</span></pre></td></tr><tr><td data-num="3"></td><td><pre> unordered_set<span class="token operator"><</span><span class="token keyword">int</span><span class="token operator">></span> ans<span class="token punctuation">;</span> <span class="token comment">// 存放结果（两数组的交集）</span></pre></td></tr><tr><td data-num="4"></td><td><pre> <span class="token keyword">for</span> <span class="token punctuation">(</span><span class="token keyword">auto</span> a <span class="token operator">:</span> nums2<span class="token punctuation">)</span></pre></td></tr><tr><td data-num="5"></td><td><pre> <span class="token keyword">if</span> <span class="token punctuation">(</span>record<span class="token punctuation">.</span><span class="token function">find</span><span class="token punctuation">(</span>a<span class="token punctuation">)</span> <span class="token operator">!=</span> record<span class="token punctuation">.</span><span class="token function">end</span><span class="token punctuation">(</span><span class="token punctuation">)</span><span class="token punctuation">)</span> <span class="token comment">//nums2 的元素 a 在 nums1 中出现过</span></pre></td></tr><tr><td data-num="6"></td><td><pre> ans<span class="token punctuation">.</span><span class="token function">insert</span><span class="token punctuation">(</span>a<span class="token punctuation">)</span><span class="token punctuation">;</span> <span class="token comment">// 将 a 添加到结果</span></pre></td></tr><tr><td data-num="7"></td><td><pre> <span class="token keyword">return</span> <span class="token generic-function"><span class="token function">vector</span><span class="token generic class-name"><span class="token operator"><</span><span class="token keyword">int</span><span class="token operator">></span></span></span><span class="token punctuation">(</span>ans<span class="token punctuation">.</span><span class="token function">begin</span><span class="token punctuation">(</span><span class="token punctuation">)</span><span class="token punctuation">,</span> ans<span class="token punctuation">.</span><span class="token function">end</span><span class="token punctuation">(</span><span class="token punctuation">)</span><span class="token punctuation">)</span><span class="token punctuation">;</span> <span class="token comment">// 返回（先拷贝到 vector ，因为返回值的类型是 vector<int>）</span></pre></td></tr><tr><td data-num="8"></td><td><pre><span class="token punctuation">}</span></pre></td></tr></tbody></table></figure><h1 id="leetcode-383-赎金信"><a class="anchor" href="#leetcode-383-赎金信">#</a> LeetCode 383. 赎金信</h1>
<p><a target="_blank" rel="noopener" href="https://leetcode-cn.com/problems/ransom-note/">LeetCode 383. Ransom Note</a></p>
<p>给你两个字符串： <code>ransomNote</code> 和 <code>magazine</code> ，判断 <code>ransomNote</code> 能不能由 <code>magazine</code> 里面的字符构成。</p>
<p>如果可以，返回 <code>true</code> ；否则返回 <code>false</code> 。</p>
<p><code>magazine</code> 中的每个字符只能在 <code>ransomNote</code> 中使用一次。</p>
<p><strong>示例 1：</strong></p>
<pre><code>输入：ransomNote = "a", magazine = "b"
输出：false
</code></pre>
<p><strong>示例 2：</strong></p>
<pre><code>输入：ransomNote = "aa", magazine = "ab"
输出：false
</code></pre>
<p><strong>示例 3：</strong></p>
<pre><code>输入：ransomNote = "aa", magazine = "aab"
输出：true
</code></pre>
<p></p>
<p><strong>提示：</strong></p>
<ul>
<li><span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mn>1</mn><mo>≤</mo></mrow><annotation encoding="application/x-tex">1 \le</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.7804em;vertical-align:-0.136em;"></span><span class="mord">1</span><span class="mspace" style="margin-right:0.2778em;"></span><span class="mrel">≤</span></span></span></span> <code>ransomNote.length</code> , <code>magazine.length</code> <span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mo>≤</mo><mn>1</mn><msup><mn>0</mn><mn>5</mn></msup></mrow><annotation encoding="application/x-tex">\le 10^5</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.7719em;vertical-align:-0.136em;"></span><span class="mrel">≤</span><span class="mspace" style="margin-right:0.2778em;"></span></span><span class="base"><span class="strut" style="height:0.8141em;"></span><span class="mord">1</span><span class="mord"><span class="mord">0</span><span class="msupsub"><span class="vlist-t"><span class="vlist-r"><span class="vlist" style="height:0.8141em;"><span style="top:-3.063em;margin-right:0.05em;"><span class="pstrut" style="height:2.7em;"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mtight">5</span></span></span></span></span></span></span></span></span></span></span></li>
<li><code>ransomNote</code> 和 <code>magazine</code> 由小写英文字母组成</li>
</ul>
<h2 id="method-哈希"><a class="anchor" href="#method-哈希">#</a> Method: 哈希</h2>
<p>由于字符串中只有小写字母，可以考虑用 数组 实现哈希法</p>
<p>解题步骤如下：</p>
<ol>
<li>
<p>定义数组 <code>count</code> ，数组元素的下标表示字符相当于 <code>a</code> 的距离，元素值表示字符在字符串 <code>magazine</code> 中出现的次数</p>
</li>
<li>
<p>依据数组 <code>count</code> 判断 <code>ransomNote</code> 的每个字符是否都在字符串 <code>magazine</code> 内</p>
<ul>
<li>由于 <code>magazine</code> 中每个字符只能在 <code>ransomNote</code> 中用一次，在遍历字符串 <code>ransomNote</code> 中的每个字符时，需要维护字符所对应的 <code>count</code> 元素值，即，“用一次就会少一次”</li>
</ul>
</li>
</ol>
<p>代码实现：</p>
<figure class="highlight cpp"><figcaption data-lang="C++"></figcaption><table><tbody><tr><td data-num="1"></td><td><pre><span class="token keyword">bool</span> <span class="token function">canConstruct</span><span class="token punctuation">(</span>string ransomNote<span class="token punctuation">,</span> string magazine<span class="token punctuation">)</span> <span class="token punctuation">{</span></pre></td></tr><tr><td data-num="2"></td><td><pre> vector<span class="token operator"><</span><span class="token keyword">int</span><span class="token operator">></span> <span class="token function">count</span><span class="token punctuation">(</span><span class="token number">26</span><span class="token punctuation">,</span> <span class="token number">0</span><span class="token punctuation">)</span><span class="token punctuation">;</span> <span class="token comment">// record the letters and their frequency in string magazine</span></pre></td></tr><tr><td data-num="3"></td><td><pre> <span class="token keyword">for</span> <span class="token punctuation">(</span><span class="token keyword">auto</span> c <span class="token operator">:</span> magazine<span class="token punctuation">)</span></pre></td></tr><tr><td data-num="4"></td><td><pre> count<span class="token punctuation">[</span>c <span class="token operator">-</span> <span class="token char">'a'</span><span class="token punctuation">]</span><span class="token operator">++</span><span class="token punctuation">;</span></pre></td></tr><tr><td data-num="5"></td><td><pre> <span class="token keyword">for</span> <span class="token punctuation">(</span><span class="token keyword">auto</span> c <span class="token operator">:</span> ransomNote<span class="token punctuation">)</span> <span class="token punctuation">{</span></pre></td></tr><tr><td data-num="6"></td><td><pre> <span class="token keyword">if</span> <span class="token punctuation">(</span>count<span class="token punctuation">[</span>c <span class="token operator">-</span> <span class="token char">'a'</span><span class="token punctuation">]</span><span class="token punctuation">)</span></pre></td></tr><tr><td data-num="7"></td><td><pre> count<span class="token punctuation">[</span>c <span class="token operator">-</span> <span class="token char">'a'</span><span class="token punctuation">]</span><span class="token operator">--</span><span class="token punctuation">;</span> <span class="token comment">// since each letter in string magazine can only be used once</span></pre></td></tr><tr><td data-num="8"></td><td><pre> <span class="token keyword">else</span></pre></td></tr><tr><td data-num="9"></td><td><pre> <span class="token keyword">return</span> <span class="token boolean">false</span><span class="token punctuation">;</span> <span class="token comment">// c cannot be found in string magazine</span></pre></td></tr><tr><td data-num="10"></td><td><pre> <span class="token punctuation">}</span></pre></td></tr><tr><td data-num="11"></td><td><pre> <span class="token keyword">return</span> <span class="token boolean">true</span><span class="token punctuation">;</span></pre></td></tr><tr><td data-num="12"></td><td><pre><span class="token punctuation">}</span></pre></td></tr></tbody></table></figure><p>时间复杂度：<span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>O</mi><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow><annotation encoding="application/x-tex">O(n)</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:1em;vertical-align:-0.25em;"></span><span class="mord mathnormal" style="margin-right:0.02778em;">O</span><span class="mopen">(</span><span class="mord mathnormal">n</span><span class="mclose">)</span></span></span></span></p>
<p>空间复杂度：<span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>O</mi><mo stretchy="false">(</mo><mn>1</mn><mo stretchy="false">)</mo></mrow><annotation encoding="application/x-tex">O(1)</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:1em;vertical-align:-0.25em;"></span><span class="mord mathnormal" style="margin-right:0.02778em;">O</span><span class="mopen">(</span><span class="mord">1</span><span class="mclose">)</span></span></span></span></p>
<blockquote>
<p>由于 <code>ransomNote</code> 和 <code>magazine</code> 都由字母组成， <code>key</code> 分布较为集中。与 set 和 map 相比，采用数组更为省时</p>
</blockquote>
<h1 id="leetcode-438-找到字符串中所有字母异位词"><a class="anchor" href="#leetcode-438-找到字符串中所有字母异位词">#</a> LeetCode 438. 找到字符串中所有字母异位词</h1>
<p><a target="_blank" rel="noopener" href="https://leetcode.cn/problems/find-all-anagrams-in-a-string/">438. Find All Anagrams in a String</a></p>
<p>给定两个字符串 <code>s</code> 和 <code>p</code> ，找到 <code>s</code> 中所有 <code>p</code> 的 <strong>异位词</strong> 的子串，返回这些子串的起始索引。不考虑答案输出的顺序。</p>
<p><strong>异位词</strong> 指由相同字母重排列形成的字符串（包括相同的字符串）。</p>
<p><strong>示例 1：</strong></p>
<pre><code>输入：s = "cbaebabacd", p = "abc"
输出：[0,6]
解释：
起始索引等于 0 的子串是 "cba", 它是 "abc" 的异位词。
起始索引等于 6 的子串是 "bac", 它是 "abc" 的异位词。
</code></pre>
<p><strong>示例 2：</strong></p>
<pre><code>输入：s = "abab", p = "ab"
输出：[0,1,2]
解释：
起始索引等于 0 的子串是 "ab", 它是 "ab" 的异位词。
起始索引等于 1 的子串是 "ba", 它是 "ab" 的异位词。
起始索引等于 2 的子串是 "ab", 它是 "ab" 的异位词。
</code></pre>
<p><strong>提示：</strong></p>
<ul>
<li><span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mn>1</mn><mo>≤</mo></mrow><annotation encoding="application/x-tex">1 \le</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.7804em;vertical-align:-0.136em;"></span><span class="mord">1</span><span class="mspace" style="margin-right:0.2778em;"></span><span class="mrel">≤</span></span></span></span> <code>s.length</code> , <code>p.length</code> <span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mo>≤</mo><mn>3</mn><mo>×</mo><mn>1</mn><msup><mn>0</mn><mn>4</mn></msup></mrow><annotation encoding="application/x-tex">\le 3 \times 10^4</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.7719em;vertical-align:-0.136em;"></span><span class="mrel">≤</span><span class="mspace" style="margin-right:0.2778em;"></span></span><span class="base"><span class="strut" style="height:0.7278em;vertical-align:-0.0833em;"></span><span class="mord">3</span><span class="mspace" style="margin-right:0.2222em;"></span><span class="mbin">×</span><span class="mspace" style="margin-right:0.2222em;"></span></span><span class="base"><span class="strut" style="height:0.8141em;"></span><span class="mord">1</span><span class="mord"><span class="mord">0</span><span class="msupsub"><span class="vlist-t"><span class="vlist-r"><span class="vlist" style="height:0.8141em;"><span style="top:-3.063em;margin-right:0.05em;"><span class="pstrut" style="height:2.7em;"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mtight">4</span></span></span></span></span></span></span></span></span></span></span></li>
<li><code>s</code> 和 <code>p</code> 仅包含小写字母</li>
</ul>
<h2 id="method-滑动窗口法"><a class="anchor" href="#method-滑动窗口法">#</a> Method: 滑动窗口法</h2>
<p>算法思路：</p>
<p>可以在字符串 s 中构造一个与字符串 p 长度相同的滑动窗口，并在滑动窗口过程中维护每种字母的数量。当每种字母在窗口中的数量等于其在字符串 p 中的数量时，窗口即为字符串 p 的异位词</p>
<p>特别地，可做以下考虑：</p>
<ul>
<li>
<p>用哈希表来统计滑动窗口与字符串 p 中每种字母的数量差。当某个字母对应的数量差为 0 时，字母在窗口中的数量等于其在字符串 p 中的数量</p>
</li>
<li>
<p>用变量 valid 来记录数量差为 0 的字母的种类数。当 valid 等于哈希表的长度时，窗口就是字符串 p 的字母异位词</p>
</li>
</ul>
<p>代码实现：</p>
<figure class="highlight cpp"><figcaption data-lang="C++"></figcaption><table><tbody><tr><td data-num="1"></td><td><pre>vector<span class="token operator"><</span><span class="token keyword">int</span><span class="token operator">></span> <span class="token function">findAnagrams</span><span class="token punctuation">(</span>string s<span class="token punctuation">,</span> string p<span class="token punctuation">)</span> <span class="token punctuation">{</span></pre></td></tr><tr><td data-num="2"></td><td><pre> vector<span class="token operator"><</span><span class="token keyword">int</span><span class="token operator">></span> ans<span class="token punctuation">;</span> <span class="token comment">// 结果数组</span></pre></td></tr><tr><td data-num="3"></td><td><pre> <span class="token keyword">int</span> sLen <span class="token operator">=</span> s<span class="token punctuation">.</span><span class="token function">size</span><span class="token punctuation">(</span><span class="token punctuation">)</span><span class="token punctuation">;</span></pre></td></tr><tr><td data-num="4"></td><td><pre> <span class="token keyword">int</span> pLen <span class="token operator">=</span> p<span class="token punctuation">.</span><span class="token function">size</span><span class="token punctuation">(</span><span class="token punctuation">)</span><span class="token punctuation">;</span></pre></td></tr><tr><td data-num="5"></td><td><pre> <span class="token keyword">if</span> <span class="token punctuation">(</span>sLen <span class="token operator"><</span> pLen<span class="token punctuation">)</span> <span class="token keyword">return</span> ans<span class="token punctuation">;</span></pre></td></tr><tr><td data-num="6"></td><td><pre></pre></td></tr><tr><td data-num="7"></td><td><pre> unordered_map<span class="token operator"><</span><span class="token keyword">char</span><span class="token punctuation">,</span> <span class="token keyword">int</span><span class="token operator">></span> record<span class="token punctuation">;</span></pre></td></tr><tr><td data-num="8"></td><td><pre> <span class="token keyword">for</span> <span class="token punctuation">(</span><span class="token keyword">auto</span> c <span class="token operator">:</span> p<span class="token punctuation">)</span></pre></td></tr><tr><td data-num="9"></td><td><pre> <span class="token operator">++</span>record<span class="token punctuation">[</span>c<span class="token punctuation">]</span><span class="token punctuation">;</span> <span class="token comment">//record [c] 表示需添加到窗口内的字符 c 的数量</span></pre></td></tr><tr><td data-num="10"></td><td><pre></pre></td></tr><tr><td data-num="11"></td><td><pre> <span class="token keyword">int</span> left <span class="token operator">=</span> <span class="token number">0</span><span class="token punctuation">;</span> <span class="token comment">// 滑动窗口的左边界</span></pre></td></tr><tr><td data-num="12"></td><td><pre> <span class="token keyword">int</span> right <span class="token operator">=</span> <span class="token number">0</span><span class="token punctuation">;</span> <span class="token comment">// 滑动窗口的右边界</span></pre></td></tr><tr><td data-num="13"></td><td><pre> <span class="token keyword">int</span> valid <span class="token operator">=</span> <span class="token number">0</span><span class="token punctuation">;</span> <span class="token comment">// 窗口内已满足数量要求的字符种类数</span></pre></td></tr><tr><td data-num="14"></td><td><pre> <span class="token keyword">while</span> <span class="token punctuation">(</span>right <span class="token operator"><</span> sLen<span class="token punctuation">)</span> <span class="token punctuation">{</span></pre></td></tr><tr><td data-num="15"></td><td><pre> <span class="token keyword">char</span> c <span class="token operator">=</span> s<span class="token punctuation">[</span>right<span class="token punctuation">]</span><span class="token punctuation">;</span></pre></td></tr><tr><td data-num="16"></td><td><pre> <span class="token comment">// 更新窗口的相关数据</span></pre></td></tr><tr><td data-num="17"></td><td><pre> <span class="token keyword">if</span> <span class="token punctuation">(</span>record<span class="token punctuation">.</span><span class="token function">count</span><span class="token punctuation">(</span>c<span class="token punctuation">)</span><span class="token punctuation">)</span> <span class="token punctuation">{</span> <span class="token comment">// 仅当 p 字符串包含的字符 c 时更新 record 和 valid</span></pre></td></tr><tr><td data-num="18"></td><td><pre> <span class="token operator">--</span>record<span class="token punctuation">[</span>c<span class="token punctuation">]</span><span class="token punctuation">;</span> <span class="token comment">//c 添加到窗口内，所需字符 c 的数量应减 1</span></pre></td></tr><tr><td data-num="19"></td><td><pre> <span class="token keyword">if</span> <span class="token punctuation">(</span>record<span class="token punctuation">[</span>c<span class="token punctuation">]</span> <span class="token operator">==</span> <span class="token number">0</span><span class="token punctuation">)</span> <span class="token comment">// 字符 c 的数量满足要求，有效字符数加 1</span></pre></td></tr><tr><td data-num="20"></td><td><pre> <span class="token operator">++</span>valid<span class="token punctuation">;</span></pre></td></tr><tr><td data-num="21"></td><td><pre> <span class="token punctuation">}</span></pre></td></tr><tr><td data-num="22"></td><td><pre> <span class="token comment">// 判断是否需要收缩窗口</span></pre></td></tr><tr><td data-num="23"></td><td><pre> <span class="token keyword">while</span> <span class="token punctuation">(</span>right <span class="token operator">-</span> left <span class="token operator">+</span> <span class="token number">1</span> <span class="token operator">></span> pLen<span class="token punctuation">)</span> <span class="token punctuation">{</span> <span class="token comment">// 窗口长度大于 pLen ，应收缩窗口</span></pre></td></tr><tr><td data-num="24"></td><td><pre> <span class="token keyword">char</span> d <span class="token operator">=</span> s<span class="token punctuation">[</span>left<span class="token punctuation">]</span><span class="token punctuation">;</span></pre></td></tr><tr><td data-num="25"></td><td><pre> <span class="token comment">// 更新窗口的相关数据</span></pre></td></tr><tr><td data-num="26"></td><td><pre> <span class="token keyword">if</span> <span class="token punctuation">(</span>record<span class="token punctuation">.</span><span class="token function">count</span><span class="token punctuation">(</span>d<span class="token punctuation">)</span><span class="token punctuation">)</span> <span class="token punctuation">{</span> <span class="token comment">// 仅当 p 字符串包含的字符 d 时更新 record 和 valid</span></pre></td></tr><tr><td data-num="27"></td><td><pre> <span class="token keyword">if</span> <span class="token punctuation">(</span>record<span class="token punctuation">[</span>d<span class="token punctuation">]</span> <span class="token operator">==</span> <span class="token number">0</span><span class="token punctuation">)</span> <span class="token comment">// 目前字符 d 满足数量要求，但其将被移出，故有效字符数减 1</span></pre></td></tr><tr><td data-num="28"></td><td><pre> <span class="token operator">--</span>valid<span class="token punctuation">;</span></pre></td></tr><tr><td data-num="29"></td><td><pre> <span class="token operator">++</span>record<span class="token punctuation">[</span>d<span class="token punctuation">]</span><span class="token punctuation">;</span> <span class="token comment">//d 从窗口内移出，所需字符 d 的数量应加 1</span></pre></td></tr><tr><td data-num="30"></td><td><pre> <span class="token punctuation">}</span></pre></td></tr><tr><td data-num="31"></td><td><pre> <span class="token operator">++</span>left<span class="token punctuation">;</span> <span class="token comment">// 窗口的左边界向右移动</span></pre></td></tr><tr><td data-num="32"></td><td><pre> <span class="token punctuation">}</span></pre></td></tr><tr><td data-num="33"></td><td><pre> <span class="token comment">// 窗口符合条件时，将起始索引加入目标数组</span></pre></td></tr><tr><td data-num="34"></td><td><pre> <span class="token keyword">if</span> <span class="token punctuation">(</span>right <span class="token operator">-</span> left <span class="token operator">+</span> <span class="token number">1</span> <span class="token operator">==</span> pLen<span class="token punctuation">)</span> <span class="token punctuation">{</span> <span class="token comment">// 窗口长度等于字符串 p 长度</span></pre></td></tr><tr><td data-num="35"></td><td><pre> <span class="token keyword">if</span> <span class="token punctuation">(</span>valid <span class="token operator">==</span> record<span class="token punctuation">.</span><span class="token function">size</span><span class="token punctuation">(</span><span class="token punctuation">)</span><span class="token punctuation">)</span> <span class="token comment">// 所有字符均满足数量要求，将索引添加到目标数组</span></pre></td></tr><tr><td data-num="36"></td><td><pre> ans<span class="token punctuation">.</span><span class="token function">push_back</span><span class="token punctuation">(</span>left<span class="token punctuation">)</span><span class="token punctuation">;</span></pre></td></tr><tr><td data-num="37"></td><td><pre> <span class="token punctuation">}</span></pre></td></tr><tr><td data-num="38"></td><td><pre> <span class="token operator">++</span>right<span class="token punctuation">;</span> <span class="token comment">// 窗口的右边界向右移动</span></pre></td></tr><tr><td data-num="39"></td><td><pre> <span class="token punctuation">}</span></pre></td></tr><tr><td data-num="40"></td><td><pre></pre></td></tr><tr><td data-num="41"></td><td><pre> <span class="token keyword">return</span> ans<span class="token punctuation">;</span></pre></td></tr><tr><td data-num="42"></td><td><pre><span class="token punctuation">}</span></pre></td></tr></tbody></table></figure><p>时间复杂度：<span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>O</mi><mo stretchy="false">(</mo><mi>n</mi><mo>+</mo><mi>m</mi><mo stretchy="false">)</mo></mrow><annotation encoding="application/x-tex">O(n + m)</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:1em;vertical-align:-0.25em;"></span><span class="mord mathnormal" style="margin-right:0.02778em;">O</span><span class="mopen">(</span><span class="mord mathnormal">n</span><span class="mspace" style="margin-right:0.2222em;"></span><span class="mbin">+</span><span class="mspace" style="margin-right:0.2222em;"></span></span><span class="base"><span class="strut" style="height:1em;vertical-align:-0.25em;"></span><span class="mord mathnormal">m</span><span class="mclose">)</span></span></span></span>，其中，<span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>n</mi></mrow><annotation encoding="application/x-tex">n</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.4306em;"></span><span class="mord mathnormal">n</span></span></span></span> 为字符串 s 的长度，<span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>m</mi></mrow><annotation encoding="application/x-tex">m</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.4306em;"></span><span class="mord mathnormal">m</span></span></span></span> 为字符串 p 的长度</p>
<p>空间复杂度：<span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>O</mi><mo stretchy="false">(</mo><mi mathvariant="normal">∣</mi><mi mathvariant="normal">Σ</mi><mi mathvariant="normal">∣</mi><mo stretchy="false">)</mo></mrow><annotation encoding="application/x-tex">O(\vert \Sigma \vert)</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:1em;vertical-align:-0.25em;"></span><span class="mord mathnormal" style="margin-right:0.02778em;">O</span><span class="mopen">(</span><span class="mord">∣Σ∣</span><span class="mclose">)</span></span></span></span>，其中，<span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi mathvariant="normal">∣</mi><mi mathvariant="normal">Σ</mi><mi mathvariant="normal">∣</mi></mrow><annotation encoding="application/x-tex">\vert \Sigma \vert</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:1em;vertical-align:-0.25em;"></span><span class="mord">∣Σ∣</span></span></span></span> 为字符串 p 中的字符种类数</p>
<p>参考：</p>
<ul>
<li><a target="_blank" rel="noopener" href="https://leetcode.cn/problems/find-all-anagrams-in-a-string/solution/hua-dong-chuang-kou-tong-yong-si-xiang-jie-jue-zi-/">labuladong</a></li>
<li><a target="_blank" rel="noopener" href="https://leetcode.cn/problems/find-all-anagrams-in-a-string/solution/zhao-dao-zi-fu-chuan-zhong-suo-you-zi-mu-xzin/">leetcode-solution</a></li>
</ul>
<h1 id="leetcode-454-四数相加ii"><a class="anchor" href="#leetcode-454-四数相加ii">#</a> LeetCode 454. 四数相加 II</h1>
<p><a target="_blank" rel="noopener" href="https://leetcode-cn.com/problems/4sum-ii/">LeetCode 454. 4Sum II</a></p>
<p>给你四个整数数组 <code>nums1</code> 、 <code>nums2</code> 、 <code>nums3</code> 和 <code>nums4</code> ，数组长度都是 <code>n</code> ，请你计算有多少个元组 <code>(i, j, k, l)</code> 能满足：</p>
<ul>
<li>
<p><code>0 <= i, j, k, l < n</code></p>
</li>
<li>
<p><code>nums1[i] + nums2[j] + nums3[k] + nums4[l] == 0</code></p>
</li>
</ul>
<p><strong>示例 1：</strong></p>
<pre><code>输入：nums1 = [1,2], nums2 = [-2,-1], nums3 = [-1,2], nums4 = [0,2]
输出：2
解释：两个元组如下：
1. (0, 0, 0, 1) -> nums1[0] + nums2[0] + nums3[0] + nums4[1] = 1 + (-2) + (-1) + 2 = 0
2. (1, 1, 0, 0) -> nums1[1] + nums2[1] + nums3[0] + nums4[0] = 2 + (-1) + (-1) + 0 = 0
</code></pre>
<p><strong>示例 2：</strong></p>
<pre><code>输入：nums1 = [0], nums2 = [0], nums3 = [0], nums4 = [0]
输出：1
</code></pre>
<p></p>
<p><strong>提示：</strong></p>
<ul>
<li><code>n == nums1.length</code></li>
<li><code>n == nums2.length</code></li>
<li><code>n == nums3.length</code></li>
<li><code>n == nums4.length</code></li>
<li><span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mn>1</mn><mo>≤</mo></mrow><annotation encoding="application/x-tex">1 \le</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.7804em;vertical-align:-0.136em;"></span><span class="mord">1</span><span class="mspace" style="margin-right:0.2778em;"></span><span class="mrel">≤</span></span></span></span> <code>n</code> <span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mo>≤</mo><mn>200</mn></mrow><annotation encoding="application/x-tex">\le 200</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.7719em;vertical-align:-0.136em;"></span><span class="mrel">≤</span><span class="mspace" style="margin-right:0.2778em;"></span></span><span class="base"><span class="strut" style="height:0.6444em;"></span><span class="mord">200</span></span></span></span></li>
<li><span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mo>−</mo><msup><mn>2</mn><mn>28</mn></msup><mo>≤</mo></mrow><annotation encoding="application/x-tex">- 2^{28} \le</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.9501em;vertical-align:-0.136em;"></span><span class="mord">−</span><span class="mord"><span class="mord">2</span><span class="msupsub"><span class="vlist-t"><span class="vlist-r"><span class="vlist" style="height:0.8141em;"><span style="top:-3.063em;margin-right:0.05em;"><span class="pstrut" style="height:2.7em;"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mtight"><span class="mord mtight">28</span></span></span></span></span></span></span></span></span><span class="mspace" style="margin-right:0.2778em;"></span><span class="mrel">≤</span></span></span></span> <code>nums1[i]</code> , <code>nums2[i]</code> , <code>nums3[i]</code> , <code>nums4[i]</code> <span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mo>≤</mo><msup><mn>2</mn><mn>28</mn></msup></mrow><annotation encoding="application/x-tex">\le 2^{28}</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.7719em;vertical-align:-0.136em;"></span><span class="mrel">≤</span><span class="mspace" style="margin-right:0.2778em;"></span></span><span class="base"><span class="strut" style="height:0.8141em;"></span><span class="mord"><span class="mord">2</span><span class="msupsub"><span class="vlist-t"><span class="vlist-r"><span class="vlist" style="height:0.8141em;"><span style="top:-3.063em;margin-right:0.05em;"><span class="pstrut" style="height:2.7em;"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mtight"><span class="mord mtight">28</span></span></span></span></span></span></span></span></span></span></span></span></li>
</ul>
<h2 id="思路-7"><a class="anchor" href="#思路-7">#</a> 思路</h2>
<p>统计 <code>nums1[i] + nums2[j] + nums3[k] + nums4[l] == 0</code> 成立的 <code>(i, j, k, l)</code> 的个数，不用考虑是否有重复的元素值相加</p>
<p>可以采用 <code>unordered_map</code> 容器记录 任意 <code>nums1[i]</code> 与 <code>nums2[j]</code> 相加的和，将其作为 <code>key</code> ，同时记录 该 <code>key</code> 值出现的次数，将其作为对应的 <code>value</code></p>
<p>然后遍历 <code>nums3[k]</code> 和 <code>nums4[l]</code> ，看 <code>unordered_map</code> 容器中是否能找到值为 <code>0 - num3[k] - num4[l]</code> 的 <code>key</code> ，将找到的所有 <code>key</code> 对应的 <code>value</code> 进行求和，即为所求</p>
<h2 id="哈希-map"><a class="anchor" href="#哈希-map">#</a> 哈希 map</h2>
<p>解题步骤：</p>
<ol>
<li>定义一个 <code>unordered_map</code> 容器，命名为 <code>map</code></li>
<li>遍历数组 <code>nums1</code> 的元素 <code>a</code> 和 数组 <code>nums2</code> 的元素 <code>b</code> ，将 <code>a + b</code> 作为哈希 <code>map</code> 的 <code>key</code> ，将值 <code>a + b</code> 出现的次数作为对应的 <code>value</code></li>
<li>定义变量 <code>count</code> ，用来记录答案</li>
<li>遍历数组 <code>nums3</code> 元素 <code>c</code> 和数组 <code>nums4</code> 元素 <code>d</code> ，如果在哈希 <code>map</code> 中找到 <code>0 - (c + d)</code> ，把对应的 <code>value</code> 累加到 <code>count</code></li>
</ol>
<p>代码实现：</p>
<figure class="highlight cpp"><figcaption data-lang="C++"></figcaption><table><tbody><tr><td data-num="1"></td><td><pre><span class="token keyword">int</span> <span class="token function">fourSumCount</span><span class="token punctuation">(</span>vector<span class="token operator"><</span><span class="token keyword">int</span><span class="token operator">></span><span class="token operator">&</span> nums1<span class="token punctuation">,</span> vector<span class="token operator"><</span><span class="token keyword">int</span><span class="token operator">></span><span class="token operator">&</span> nums2<span class="token punctuation">,</span> vector<span class="token operator"><</span><span class="token keyword">int</span><span class="token operator">></span><span class="token operator">&</span> nums3<span class="token punctuation">,</span> vector<span class="token operator"><</span><span class="token keyword">int</span><span class="token operator">></span><span class="token operator">&</span> nums4<span class="token punctuation">)</span> <span class="token punctuation">{</span></pre></td></tr><tr><td data-num="2"></td><td><pre> <span class="token comment">// 统计 nums1 和 nums2 数组中任意两元素之和</span></pre></td></tr><tr><td data-num="3"></td><td><pre> unordered_map<span class="token operator"><</span><span class="token keyword">int</span><span class="token punctuation">,</span> <span class="token keyword">int</span><span class="token operator">></span> map<span class="token punctuation">;</span></pre></td></tr><tr><td data-num="4"></td><td><pre> <span class="token keyword">for</span> <span class="token punctuation">(</span><span class="token keyword">auto</span> a <span class="token operator">:</span> nums1<span class="token punctuation">)</span></pre></td></tr><tr><td data-num="5"></td><td><pre> <span class="token keyword">for</span> <span class="token punctuation">(</span><span class="token keyword">auto</span> b <span class="token operator">:</span> nums2<span class="token punctuation">)</span></pre></td></tr><tr><td data-num="6"></td><td><pre> map<span class="token punctuation">[</span>a <span class="token operator">+</span> b<span class="token punctuation">]</span><span class="token operator">++</span><span class="token punctuation">;</span></pre></td></tr><tr><td data-num="7"></td><td><pre> <span class="token comment">// 统计 a + b + c + d = 0 的次数</span></pre></td></tr><tr><td data-num="8"></td><td><pre> <span class="token keyword">int</span> count <span class="token operator">=</span> <span class="token number">0</span><span class="token punctuation">;</span></pre></td></tr><tr><td data-num="9"></td><td><pre> <span class="token keyword">for</span> <span class="token punctuation">(</span><span class="token keyword">auto</span> c <span class="token operator">:</span> nums3<span class="token punctuation">)</span></pre></td></tr><tr><td data-num="10"></td><td><pre> <span class="token keyword">for</span> <span class="token punctuation">(</span><span class="token keyword">auto</span> d <span class="token operator">:</span> nums4<span class="token punctuation">)</span></pre></td></tr><tr><td data-num="11"></td><td><pre> <span class="token keyword">if</span> <span class="token punctuation">(</span>map<span class="token punctuation">.</span><span class="token function">find</span><span class="token punctuation">(</span><span class="token number">0</span> <span class="token operator">-</span> c <span class="token operator">-</span> d<span class="token punctuation">)</span> <span class="token operator">!=</span> map<span class="token punctuation">.</span><span class="token function">end</span><span class="token punctuation">(</span><span class="token punctuation">)</span><span class="token punctuation">)</span></pre></td></tr><tr><td data-num="12"></td><td><pre> count <span class="token operator">+=</span> map<span class="token punctuation">[</span><span class="token number">0</span> <span class="token operator">-</span> c <span class="token operator">-</span> d<span class="token punctuation">]</span><span class="token punctuation">;</span></pre></td></tr><tr><td data-num="13"></td><td><pre> <span class="token keyword">return</span> count<span class="token punctuation">;</span></pre></td></tr><tr><td data-num="14"></td><td><pre><span class="token punctuation">}</span></pre></td></tr></tbody></table></figure><p>时间复杂度：<span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>O</mi><mo stretchy="false">(</mo><msup><mi>n</mi><mn>2</mn></msup><mo stretchy="false">)</mo></mrow><annotation encoding="application/x-tex">O(n^2)</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:1.0641em;vertical-align:-0.25em;"></span><span class="mord mathnormal" style="margin-right:0.02778em;">O</span><span class="mopen">(</span><span class="mord"><span class="mord mathnormal">n</span><span class="msupsub"><span class="vlist-t"><span class="vlist-r"><span class="vlist" style="height:0.8141em;"><span style="top:-3.063em;margin-right:0.05em;"><span class="pstrut" style="height:2.7em;"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mtight">2</span></span></span></span></span></span></span></span><span class="mclose">)</span></span></span></span></p>
<ul>
<li>两个 <code>for</code> 循环，时间复杂度为 <span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>O</mi><mo stretchy="false">(</mo><msup><mi>n</mi><mn>2</mn></msup><mo stretchy="false">)</mo></mrow><annotation encoding="application/x-tex">O(n^2)</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:1.0641em;vertical-align:-0.25em;"></span><span class="mord mathnormal" style="margin-right:0.02778em;">O</span><span class="mopen">(</span><span class="mord"><span class="mord mathnormal">n</span><span class="msupsub"><span class="vlist-t"><span class="vlist-r"><span class="vlist" style="height:0.8141em;"><span style="top:-3.063em;margin-right:0.05em;"><span class="pstrut" style="height:2.7em;"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mtight">2</span></span></span></span></span></span></span></span><span class="mclose">)</span></span></span></span></li>
<li>每次哈希查找的时间复杂度为 <span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>O</mi><mo stretchy="false">(</mo><mn>1</mn><mo stretchy="false">)</mo></mrow><annotation encoding="application/x-tex">O(1)</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:1em;vertical-align:-0.25em;"></span><span class="mord mathnormal" style="margin-right:0.02778em;">O</span><span class="mopen">(</span><span class="mord">1</span><span class="mclose">)</span></span></span></span></li>
</ul>
<p>空间复杂度：<span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>O</mi><mo stretchy="false">(</mo><msup><mi>n</mi><mn>2</mn></msup><mo stretchy="false">)</mo></mrow><annotation encoding="application/x-tex">O(n^2)</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:1.0641em;vertical-align:-0.25em;"></span><span class="mord mathnormal" style="margin-right:0.02778em;">O</span><span class="mopen">(</span><span class="mord"><span class="mord mathnormal">n</span><span class="msupsub"><span class="vlist-t"><span class="vlist-r"><span class="vlist" style="height:0.8141em;"><span style="top:-3.063em;margin-right:0.05em;"><span class="pstrut" style="height:2.7em;"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mtight">2</span></span></span></span></span></span></span></span><span class="mclose">)</span></span></span></span>，哈希 <code>map</code> 所需要的空间</p>
<p>参考：<a target="_blank" rel="noopener" href="https://www.programmercarl.com/0454.%E5%9B%9B%E6%95%B0%E7%9B%B8%E5%8A%A0II.html#%E5%85%B6%E4%BB%96%E8%AF%AD%E8%A8%80%E7%89%88%E6%9C%AC">代码随想录：四数相加 II</a></p>
<h1 id="leetcode-49-字母异位词分组"><a class="anchor" href="#leetcode-49-字母异位词分组">#</a> LeetCode 49. 字母异位词分组</h1>
<p><a target="_blank" rel="noopener" href="https://leetcode.cn/problems/group-anagrams/">49. Group Anagrams</a></p>
<p>给你一个字符串数组，请你将 <strong>字母异位词</strong> 组合在一起。可以按任意顺序返回结果列表。</p>
<p><strong>字母异位词</strong> 是由重新排列源单词的字母得到的一个新单词，所有源单词中的字母通常恰好只用一次。</p>
<p><strong>示例 1：</strong></p>
<pre><code>输入：strs = ["eat","tea","tan","ate","nat","bat"]
输出：[["bat"],["nat","tan"],["ate","eat","tea"]]
</code></pre>
<p><strong>示例 2：</strong></p>
<pre><code>输入：strs = [""]
输出：[[""]]
</code></pre>
<p><strong>示例 3：</strong></p>
<pre><code>输入：strs = ["a"]
输出：[["a"]]
</code></pre>
<p><strong>提示：</strong></p>
<ul>
<li><span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mn>1</mn><mo>≤</mo></mrow><annotation encoding="application/x-tex">1 \le</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.7804em;vertical-align:-0.136em;"></span><span class="mord">1</span><span class="mspace" style="margin-right:0.2778em;"></span><span class="mrel">≤</span></span></span></span> <code>strs.length</code> <span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mo>≤</mo><mn>1</mn><msup><mn>0</mn><mn>4</mn></msup></mrow><annotation encoding="application/x-tex">\le 10^4</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.7719em;vertical-align:-0.136em;"></span><span class="mrel">≤</span><span class="mspace" style="margin-right:0.2778em;"></span></span><span class="base"><span class="strut" style="height:0.8141em;"></span><span class="mord">1</span><span class="mord"><span class="mord">0</span><span class="msupsub"><span class="vlist-t"><span class="vlist-r"><span class="vlist" style="height:0.8141em;"><span style="top:-3.063em;margin-right:0.05em;"><span class="pstrut" style="height:2.7em;"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mtight">4</span></span></span></span></span></span></span></span></span></span></span></li>
<li><span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mn>0</mn><mo>≤</mo></mrow><annotation encoding="application/x-tex">0 \le</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.7804em;vertical-align:-0.136em;"></span><span class="mord">0</span><span class="mspace" style="margin-right:0.2778em;"></span><span class="mrel">≤</span></span></span></span> <code>strs[i].length</code> <span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mo>≤</mo><mn>100</mn></mrow><annotation encoding="application/x-tex">\le 100</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.7719em;vertical-align:-0.136em;"></span><span class="mrel">≤</span><span class="mspace" style="margin-right:0.2778em;"></span></span><span class="base"><span class="strut" style="height:0.6444em;"></span><span class="mord">100</span></span></span></span></li>
<li><code>strs[i]</code> 仅包含小写字母</li>
</ul>
<h2 id="method-1-排序-哈希"><a class="anchor" href="#method-1-排序-哈希">#</a> Method 1: 排序 + 哈希</h2>
<p>算法思路：</p>
<p>两个字符串互为字母异位词，当且仅当它们包含的字母相同，因此，对两个字母异位词分别进行排序，所得的字符串一定相同</p>
<p>可以将排序之后的字符串作为哈希表的键，哈希表的值存放每一组字母异位词</p>
<p>代码实现：</p>
<figure class="highlight cpp"><figcaption data-lang="C++"></figcaption><table><tbody><tr><td data-num="1"></td><td><pre>vector<span class="token operator"><</span>vector<span class="token operator"><</span>string<span class="token operator">>></span> <span class="token function">groupAnagrams</span><span class="token punctuation">(</span>vector<span class="token operator"><</span>string<span class="token operator">></span><span class="token operator">&</span> strs<span class="token punctuation">)</span> <span class="token punctuation">{</span></pre></td></tr><tr><td data-num="2"></td><td><pre> unordered_map<span class="token operator"><</span>string<span class="token punctuation">,</span> vector<span class="token operator"><</span>string<span class="token operator">>></span> hash<span class="token punctuation">;</span></pre></td></tr><tr><td data-num="3"></td><td><pre> <span class="token keyword">for</span> <span class="token punctuation">(</span><span class="token keyword">auto</span> <span class="token operator">&</span>s <span class="token operator">:</span> strs<span class="token punctuation">)</span> <span class="token punctuation">{</span></pre></td></tr><tr><td data-num="4"></td><td><pre> string key <span class="token operator">=</span> s<span class="token punctuation">;</span></pre></td></tr><tr><td data-num="5"></td><td><pre> <span class="token function">sort</span><span class="token punctuation">(</span>key<span class="token punctuation">.</span><span class="token function">begin</span><span class="token punctuation">(</span><span class="token punctuation">)</span><span class="token punctuation">,</span> key<span class="token punctuation">.</span><span class="token function">end</span><span class="token punctuation">(</span><span class="token punctuation">)</span><span class="token punctuation">)</span><span class="token punctuation">;</span></pre></td></tr><tr><td data-num="6"></td><td><pre> hash<span class="token punctuation">[</span>key<span class="token punctuation">]</span><span class="token punctuation">.</span><span class="token function">push_back</span><span class="token punctuation">(</span>s<span class="token punctuation">)</span><span class="token punctuation">;</span></pre></td></tr><tr><td data-num="7"></td><td><pre> <span class="token punctuation">}</span></pre></td></tr><tr><td data-num="8"></td><td><pre> vector<span class="token operator"><</span>vector<span class="token operator"><</span>string<span class="token operator">>></span> ans<span class="token punctuation">;</span></pre></td></tr><tr><td data-num="9"></td><td><pre> <span class="token keyword">for</span> <span class="token punctuation">(</span><span class="token keyword">auto</span> it <span class="token operator">=</span> hash<span class="token punctuation">.</span><span class="token function">begin</span><span class="token punctuation">(</span><span class="token punctuation">)</span><span class="token punctuation">;</span> it <span class="token operator">!=</span> hash<span class="token punctuation">.</span><span class="token function">end</span><span class="token punctuation">(</span><span class="token punctuation">)</span><span class="token punctuation">;</span> it<span class="token operator">++</span><span class="token punctuation">)</span> <span class="token punctuation">{</span></pre></td></tr><tr><td data-num="10"></td><td><pre> ans<span class="token punctuation">.</span><span class="token function">push_back</span><span class="token punctuation">(</span>it<span class="token operator">-></span>second<span class="token punctuation">)</span><span class="token punctuation">;</span></pre></td></tr><tr><td data-num="11"></td><td><pre> <span class="token punctuation">}</span></pre></td></tr><tr><td data-num="12"></td><td><pre> <span class="token keyword">return</span> ans<span class="token punctuation">;</span></pre></td></tr><tr><td data-num="13"></td><td><pre><span class="token punctuation">}</span></pre></td></tr></tbody></table></figure><p>时间复杂度：<span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>O</mi><mo stretchy="false">(</mo><mi>n</mi><mi>k</mi><mi>log</mi><mo></mo><mi>k</mi><mo stretchy="false">)</mo></mrow><annotation encoding="application/x-tex">O(n k \log{k})</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:1em;vertical-align:-0.25em;"></span><span class="mord mathnormal" style="margin-right:0.02778em;">O</span><span class="mopen">(</span><span class="mord mathnormal" style="margin-right:0.03148em;">nk</span><span class="mspace" style="margin-right:0.1667em;"></span><span class="mop">lo<span style="margin-right:0.01389em;">g</span></span><span class="mspace" style="margin-right:0.1667em;"></span><span class="mord"><span class="mord mathnormal" style="margin-right:0.03148em;">k</span></span><span class="mclose">)</span></span></span></span>，其中，<span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>n</mi></mrow><annotation encoding="application/x-tex">n</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.4306em;"></span><span class="mord mathnormal">n</span></span></span></span> 是 <code>strs</code> 的长度，<span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>k</mi></mrow><annotation encoding="application/x-tex">k</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.6944em;"></span><span class="mord mathnormal" style="margin-right:0.03148em;">k</span></span></span></span> 是 <code>strs</code> 中字符串的最大长度</p>
<ul>
<li>遍历 <code>strs</code> 的时间复杂度为 <span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>O</mi><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow><annotation encoding="application/x-tex">O(n)</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:1em;vertical-align:-0.25em;"></span><span class="mord mathnormal" style="margin-right:0.02778em;">O</span><span class="mopen">(</span><span class="mord mathnormal">n</span><span class="mclose">)</span></span></span></span></li>
<li>对 <code>strs</code> 中的每个字符串进行排序，时间复杂度为 <span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>O</mi><mo stretchy="false">(</mo><mi>k</mi><mi>log</mi><mo></mo><mi>k</mi><mo stretchy="false">)</mo></mrow><annotation encoding="application/x-tex">O(k \log{k})</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:1em;vertical-align:-0.25em;"></span><span class="mord mathnormal" style="margin-right:0.02778em;">O</span><span class="mopen">(</span><span class="mord mathnormal" style="margin-right:0.03148em;">k</span><span class="mspace" style="margin-right:0.1667em;"></span><span class="mop">lo<span style="margin-right:0.01389em;">g</span></span><span class="mspace" style="margin-right:0.1667em;"></span><span class="mord"><span class="mord mathnormal" style="margin-right:0.03148em;">k</span></span><span class="mclose">)</span></span></span></span></li>
<li>哈希查找的时间复杂度为 <span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>O</mi><mo stretchy="false">(</mo><mn>1</mn><mo stretchy="false">)</mo></mrow><annotation encoding="application/x-tex">O(1)</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:1em;vertical-align:-0.25em;"></span><span class="mord mathnormal" style="margin-right:0.02778em;">O</span><span class="mopen">(</span><span class="mord">1</span><span class="mclose">)</span></span></span></span></li>
</ul>
<p>空间复杂度：<span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>O</mi><mo stretchy="false">(</mo><mi>n</mi><mi>k</mi><mo stretchy="false">)</mo></mrow><annotation encoding="application/x-tex">O(n k)</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:1em;vertical-align:-0.25em;"></span><span class="mord mathnormal" style="margin-right:0.02778em;">O</span><span class="mopen">(</span><span class="mord mathnormal" style="margin-right:0.03148em;">nk</span><span class="mclose">)</span></span></span></span>，需要用哈希表存储所有字母异位词</p>
<h2 id="method-2-计数-哈希"><a class="anchor" href="#method-2-计数-哈希">#</a> Method 2: 计数 + 哈希</h2>
<p>算法思路：</p>
<p>两个字母异位词中的每一种字母的出现次数一定相同</p>
<p>可以使用字符串表示字母的出现次数，将其作为哈希表的键</p>
<p>参考：<a target="_blank" rel="noopener" href="https://leetcode.cn/problems/group-anagrams/solution/zi-mu-yi-wei-ci-fen-zu-by-leetcode-solut-gyoc/">leetcode-solution</a></p>
<h1 id="leetcode-554-砖墙"><a class="anchor" href="#leetcode-554-砖墙">#</a> LeetCode 554. 砖墙</h1>
<p>你的面前有一堵矩形的、由 <code>n</code> 行砖块组成的砖墙。这些砖块高度相同（也就是一个单位高）但是宽度不同。每一行砖块的宽度之和相等。</p>
<p>你现在要画一条 <strong>自顶向下</strong> 的、穿过 <strong>最少</strong> 砖块的垂线。如果你画的线只是从砖块的边缘经过，就不算穿过这块砖。<strong>你不能沿着墙的两个垂直边缘之一画线，这样显然是没有穿过一块砖的</strong> 。</p>
<p>给你一个二维数组 <code>wall</code> ，该数组包含这堵墙的相关信息。其中， <code>wall[i]</code> 是一个代表从左至右每块砖的宽度的数组。你需要找出怎样画才能使这条线 <strong>穿过的砖块数量最少</strong> ，并且返回 <strong>穿过的砖块数量</strong> 。</p>
<p><img loading="lazy" data-src="LeetCode-%E5%93%88%E5%B8%8C%E8%A1%A8%E4%B8%93%E9%A2%98/LeetCode554_Example.webp" alt=""></p>
<p><strong>示例 1：</strong></p>
<pre><code>输入：wall = [[1,2,2,1],[3,1,2],[1,3,2],[2,4],[3,1,2],[1,3,1,1]]
输出：2
</code></pre>
<p><strong>示例 2：</strong></p>
<pre><code>输入：wall = [[1],[1],[1]]
输出：3
</code></pre>
<p><strong>提示：</strong></p>
<ul>
<li><code>n == wall.length</code></li>
<li><span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mn>1</mn><mo>≤</mo></mrow><annotation encoding="application/x-tex">1 \le</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.7804em;vertical-align:-0.136em;"></span><span class="mord">1</span><span class="mspace" style="margin-right:0.2778em;"></span><span class="mrel">≤</span></span></span></span> <code>n</code> <span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mo>≤</mo><mn>1</mn><msup><mn>0</mn><mn>4</mn></msup></mrow><annotation encoding="application/x-tex">\le 10^4</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.7719em;vertical-align:-0.136em;"></span><span class="mrel">≤</span><span class="mspace" style="margin-right:0.2778em;"></span></span><span class="base"><span class="strut" style="height:0.8141em;"></span><span class="mord">1</span><span class="mord"><span class="mord">0</span><span class="msupsub"><span class="vlist-t"><span class="vlist-r"><span class="vlist" style="height:0.8141em;"><span style="top:-3.063em;margin-right:0.05em;"><span class="pstrut" style="height:2.7em;"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mtight">4</span></span></span></span></span></span></span></span></span></span></span></li>
<li><span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mn>1</mn><mo>≤</mo></mrow><annotation encoding="application/x-tex">1 \le</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.7804em;vertical-align:-0.136em;"></span><span class="mord">1</span><span class="mspace" style="margin-right:0.2778em;"></span><span class="mrel">≤</span></span></span></span> <code>wall[i].length</code> <span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mo>≤</mo><mn>1</mn><msup><mn>0</mn><mn>4</mn></msup></mrow><annotation encoding="application/x-tex">\le 10^4</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.7719em;vertical-align:-0.136em;"></span><span class="mrel">≤</span><span class="mspace" style="margin-right:0.2778em;"></span></span><span class="base"><span class="strut" style="height:0.8141em;"></span><span class="mord">1</span><span class="mord"><span class="mord">0</span><span class="msupsub"><span class="vlist-t"><span class="vlist-r"><span class="vlist" style="height:0.8141em;"><span style="top:-3.063em;margin-right:0.05em;"><span class="pstrut" style="height:2.7em;"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mtight">4</span></span></span></span></span></span></span></span></span></span></span></li>
<li><span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mn>1</mn><mo>≤</mo></mrow><annotation encoding="application/x-tex">1 \le</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.7804em;vertical-align:-0.136em;"></span><span class="mord">1</span><span class="mspace" style="margin-right:0.2778em;"></span><span class="mrel">≤</span></span></span></span> <code>sum(wall[i].length)</code> <span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mo>≤</mo><mn>2</mn><mo>×</mo><mn>1</mn><msup><mn>0</mn><mn>4</mn></msup></mrow><annotation encoding="application/x-tex">\le 2 \times 10^4</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.7719em;vertical-align:-0.136em;"></span><span class="mrel">≤</span><span class="mspace" style="margin-right:0.2778em;"></span></span><span class="base"><span class="strut" style="height:0.7278em;vertical-align:-0.0833em;"></span><span class="mord">2</span><span class="mspace" style="margin-right:0.2222em;"></span><span class="mbin">×</span><span class="mspace" style="margin-right:0.2222em;"></span></span><span class="base"><span class="strut" style="height:0.8141em;"></span><span class="mord">1</span><span class="mord"><span class="mord">0</span><span class="msupsub"><span class="vlist-t"><span class="vlist-r"><span class="vlist" style="height:0.8141em;"><span style="top:-3.063em;margin-right:0.05em;"><span class="pstrut" style="height:2.7em;"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mtight">4</span></span></span></span></span></span></span></span></span></span></span></li>
<li>对于每一行 <code>i</code> ， <code>sum(wall[i])</code> 是相同的</li>
<li><span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mn>1</mn><mo>≤</mo></mrow><annotation encoding="application/x-tex">1 \le</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.7804em;vertical-align:-0.136em;"></span><span class="mord">1</span><span class="mspace" style="margin-right:0.2778em;"></span><span class="mrel">≤</span></span></span></span> <code>wall[i][j]</code> <span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mo>≤</mo><msup><mn>2</mn><mn>31</mn></msup><mo>−</mo><mn>1</mn></mrow><annotation encoding="application/x-tex">\le 2^{31} - 1</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.7719em;vertical-align:-0.136em;"></span><span class="mrel">≤</span><span class="mspace" style="margin-right:0.2778em;"></span></span><span class="base"><span class="strut" style="height:0.8974em;vertical-align:-0.0833em;"></span><span class="mord"><span class="mord">2</span><span class="msupsub"><span class="vlist-t"><span class="vlist-r"><span class="vlist" style="height:0.8141em;"><span style="top:-3.063em;margin-right:0.05em;"><span class="pstrut" style="height:2.7em;"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mtight"><span class="mord mtight">31</span></span></span></span></span></span></span></span></span><span class="mspace" style="margin-right:0.2222em;"></span><span class="mbin">−</span><span class="mspace" style="margin-right:0.2222em;"></span></span><span class="base"><span class="strut" style="height:0.6444em;"></span><span class="mord">1</span></span></span></span></li>
</ul>
<h2 id="method-哈希-2"><a class="anchor" href="#method-哈希-2">#</a> Method: 哈希</h2>
<h2 id="算法思路-2"><a class="anchor" href="#算法思路-2">#</a> 算法思路</h2>
<p>穿过最少的砖块 等价于 穿过最多的间隙</p>
<p>可以使用哈希表记录每个间隙的出现次数（以间隙索引位置为 key ，以间隙出现次数为 value），然后从所有行中找出间隙最大出现次数，利用 <strong>行数 减去 间隙出现最大次数 即为 穿过砖块最小数</strong></p>
<p>其中，每一行中的间隙的索引 需要按 <strong>前缀和</strong> 方法求得，例如，第一行的第一个间隙的索引为 <code>wall[0][0]</code> ，第二个间隙的索引为 <code>wall[0][0] + wall[0][1]</code> ，第 i 个间隙的索引为 <code>wall[0][0] + ... + wall[0][i]</code></p>
<p>注意：不能沿着砖墙两侧的最边缘画线，因此不需要统计砖墙两侧的间隙</p>
<p>如下图所示，间隙 4 在所有行中出现次数最多，出现次数为 4 次，而总行数为 6 ，因此穿过砖块数为 2</p>
<p><img loading="lazy" data-src="LeetCode-%E5%93%88%E5%B8%8C%E8%A1%A8%E4%B8%93%E9%A2%98/LeetCode554_Flow.webp" alt="" height="150px"></p>
<h2 id="代码实现-2"><a class="anchor" href="#代码实现-2">#</a> 代码实现</h2>
<figure class="highlight cpp"><figcaption data-lang="C++"></figcaption><table><tbody><tr><td data-num="1"></td><td><pre><span class="token keyword">int</span> <span class="token function">leastBricks</span><span class="token punctuation">(</span>vector<span class="token operator"><</span>vector<span class="token operator"><</span><span class="token keyword">int</span><span class="token operator">>></span><span class="token operator">&</span> wall<span class="token punctuation">)</span> <span class="token punctuation">{</span></pre></td></tr><tr><td data-num="2"></td><td><pre> unordered_map<span class="token operator"><</span><span class="token keyword">int</span><span class="token punctuation">,</span> <span class="token keyword">int</span><span class="token operator">></span> hash<span class="token punctuation">;</span></pre></td></tr><tr><td data-num="3"></td><td><pre> <span class="token keyword">for</span> <span class="token punctuation">(</span><span class="token keyword">int</span> i <span class="token operator">=</span> <span class="token number">0</span><span class="token punctuation">;</span> i <span class="token operator"><</span> wall<span class="token punctuation">.</span><span class="token function">size</span><span class="token punctuation">(</span><span class="token punctuation">)</span><span class="token punctuation">;</span> <span class="token operator">++</span>i<span class="token punctuation">)</span> <span class="token punctuation">{</span></pre></td></tr><tr><td data-num="4"></td><td><pre> <span class="token keyword">int</span> sum <span class="token operator">=</span> <span class="token number">0</span><span class="token punctuation">;</span></pre></td></tr><tr><td data-num="5"></td><td><pre> <span class="token keyword">for</span> <span class="token punctuation">(</span><span class="token keyword">int</span> j <span class="token operator">=</span> <span class="token number">0</span><span class="token punctuation">;</span> j <span class="token operator"><</span> wall<span class="token punctuation">[</span>i<span class="token punctuation">]</span><span class="token punctuation">.</span><span class="token function">size</span><span class="token punctuation">(</span><span class="token punctuation">)</span> <span class="token operator">-</span> <span class="token number">1</span><span class="token punctuation">;</span> <span class="token operator">++</span>j<span class="token punctuation">)</span> <span class="token punctuation">{</span> <span class="token comment">// 不能沿着最边缘画线</span></pre></td></tr><tr><td data-num="6"></td><td><pre> sum <span class="token operator">+=</span> wall<span class="token punctuation">[</span>i<span class="token punctuation">]</span><span class="token punctuation">[</span>j<span class="token punctuation">]</span><span class="token punctuation">;</span> <span class="token comment">// 间隙的索引</span></pre></td></tr><tr><td data-num="7"></td><td><pre> <span class="token operator">++</span>hash<span class="token punctuation">[</span>sum<span class="token punctuation">]</span><span class="token punctuation">;</span></pre></td></tr><tr><td data-num="8"></td><td><pre> <span class="token punctuation">}</span></pre></td></tr><tr><td data-num="9"></td><td><pre> <span class="token punctuation">}</span></pre></td></tr><tr><td data-num="10"></td><td><pre> <span class="token keyword">int</span> maxCnt <span class="token operator">=</span> <span class="token number">0</span><span class="token punctuation">;</span></pre></td></tr><tr><td data-num="11"></td><td><pre> <span class="token keyword">for</span> <span class="token punctuation">(</span><span class="token keyword">auto</span> it <span class="token operator">:</span> hash<span class="token punctuation">)</span> <span class="token punctuation">{</span></pre></td></tr><tr><td data-num="12"></td><td><pre> maxCnt <span class="token operator">=</span> <span class="token function">max</span><span class="token punctuation">(</span>maxCnt<span class="token punctuation">,</span> it<span class="token punctuation">.</span>second<span class="token punctuation">)</span><span class="token punctuation">;</span></pre></td></tr><tr><td data-num="13"></td><td><pre> <span class="token punctuation">}</span></pre></td></tr><tr><td data-num="14"></td><td><pre> <span class="token keyword">return</span> wall<span class="token punctuation">.</span><span class="token function">size</span><span class="token punctuation">(</span><span class="token punctuation">)</span> <span class="token operator">-</span> maxCnt<span class="token punctuation">;</span></pre></td></tr><tr><td data-num="15"></td><td><pre><span class="token punctuation">}</span></pre></td></tr></tbody></table></figure><h2 id="复杂度分析-2"><a class="anchor" href="#复杂度分析-2">#</a> 复杂度分析</h2>
<p>时间复杂度：<span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>O</mi><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow><annotation encoding="application/x-tex">O(n)</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:1em;vertical-align:-0.25em;"></span><span class="mord mathnormal" style="margin-right:0.02778em;">O</span><span class="mopen">(</span><span class="mord mathnormal">n</span><span class="mclose">)</span></span></span></span>，其中 <span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>n</mi></mrow><annotation encoding="application/x-tex">n</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.4306em;"></span><span class="mord mathnormal">n</span></span></span></span> 为砖块总个数</p>
<p>空间复杂度：<span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>O</mi><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow><annotation encoding="application/x-tex">O(n)</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:1em;vertical-align:-0.25em;"></span><span class="mord mathnormal" style="margin-right:0.02778em;">O</span><span class="mopen">(</span><span class="mord mathnormal">n</span><span class="mclose">)</span></span></span></span></p>
<h2 id="参考资料-2"><a class="anchor" href="#参考资料-2">#</a> 参考资料</h2>
<ul>
<li><a target="_blank" rel="noopener" href="https://leetcode.cn/problems/brick-wall/solution/gong-shui-san-xie-zheng-nan-ze-fan-shi-y-gsri/">宫水三叶：使用哈希表求解</a></li>
<li><a target="_blank" rel="noopener" href="https://leetcode.cn/problems/brick-wall/solution/zhuan-qiang-by-leetcode-solution-2kls/">力扣官方题解</a></li>
</ul>
<div class="tags"><a href="/tags/%E5%93%88%E5%B8%8C%E8%A1%A8/" rel="tag"><i class="ic i-tag"></i>哈希表</a></div></div><footer><div class="meta"><span class="item"><span class="icon"><i class="ic i-calendar-check"></i></span><span class="text">更新于</span><time title="修改时间：2024-06-08 23:08:30" itemprop="dateModified" datetime="2024-06-08T23:08:30+08:00">2024-06-08</time></span></div><div id="copyright"><ul><li class="author"><strong>本文作者：</strong>Jiankychen<i class="ic i-at"><em>@</em></i>Jiankychen's Blog</li><li class="link"><strong>本文链接：</strong><a href="https://jiankychen.github.io/leetcode-hashtable.html" title="LeetCode - 哈希表专题">https://jiankychen.github.io/leetcode-hashtable.html</a></li><li class="license"><strong>版权声明：</strong>本站所有文章除特别声明外，均采用 <a target="_blank" rel="noopener" href="https://creativecommons.org/licenses/by-nc-sa/4.0/deed.zh"><i class="ic i-creative-commons"><em>(CC)</em></i>BY-NC-SA</a> 许可协议。转载请注明出处！</li></ul></div></footer></article></div><div class="post-nav"><div class="item left"><a href="/graphics-theory.html" rel="prev" itemprop="url" data-background-image="https://img.timelessq.com/images/2022/07/26/99fb5ff897a82984470abf5e2a235d94.jpg" title="图"><span class="type">上一篇</span><span class="category"><i class="ic i-flag"></i>Data Structure</span><h3>图</h3></a></div><div class="item right"><a href="/leetcode-string.html" rel="next" itemprop="url" data-background-image="https://img.timelessq.com/images/2022/07/26/488297bfd0233b6c6a444f1860e55d45.jpg" title="LeetCode - 字符串专题"><span class="type">下一篇</span><span class="category"><i class="ic i-flag"></i>Coding</span><h3>LeetCode - 字符串专题</h3></a></div></div><div class="wrap" id="comments"></div></div><div id="sidebar"><div class="inner"><div class="panels"><div class="inner"><div class="contents panel pjax" data-title="文章目录"><ol class="toc"><li class="toc-item toc-level-1"><a class="toc-link" href="#leetcode-1-%E4%B8%A4%E6%95%B0%E4%B9%8B%E5%92%8C"><span class="toc-number">1.</span> <span class="toc-text"> LeetCode 1. 两数之和</span></a><ol class="toc-child"><li class="toc-item toc-level-3"><a class="toc-link" href="#%E6%80%9D%E8%B7%AF"><span class="toc-number">1.0.1.</span> <span class="toc-text"> 思路</span></a></li></ol></li><li class="toc-item toc-level-2"><a class="toc-link" href="#method-%E5%93%88%E5%B8%8C-map"><span class="toc-number">1.1.</span> <span class="toc-text"> Method: 哈希 map</span></a></li></ol><li class="toc-item toc-level-1"><a class="toc-link" href="#leetcode-128-%E6%9C%80%E9%95%BF%E8%BF%9E%E7%BB%AD%E5%BA%8F%E5%88%97"><span class="toc-number">2.</span> <span class="toc-text"> LeetCode 128. 最长连续序列</span></a><ol class="toc-child"><li class="toc-item toc-level-2"><a class="toc-link" href="#method-%E5%93%88%E5%B8%8C%E8%A1%A8"><span class="toc-number">2.1.</span> <span class="toc-text"> Method: 哈希表</span></a></li></ol></li><li class="toc-item toc-level-1"><a class="toc-link" href="#leetcode-1296-%E5%88%92%E5%88%86%E6%95%B0%E7%BB%84%E4%B8%BA%E8%BF%9E%E7%BB%AD%E6%95%B0%E5%AD%97%E7%9A%84%E9%9B%86%E5%90%88"><span class="toc-number">3.</span> <span class="toc-text"> LeetCode 1296. 划分数组为连续数字的集合</span></a><ol class="toc-child"><li class="toc-item toc-level-2"><a class="toc-link" href="#method-%E6%8E%92%E5%BA%8F-%E5%93%88%E5%B8%8C"><span class="toc-number">3.1.</span> <span class="toc-text"> Method: 排序 + 哈希</span></a></li><li class="toc-item toc-level-2"><a class="toc-link" href="#%E7%AE%97%E6%B3%95%E6%80%9D%E8%B7%AF"><span class="toc-number">3.2.</span> <span class="toc-text"> 算法思路</span></a></li><li class="toc-item toc-level-2"><a class="toc-link" href="#%E4%BB%A3%E7%A0%81%E5%AE%9E%E7%8E%B0"><span class="toc-number">3.3.</span> <span class="toc-text"> 代码实现</span></a></li><li class="toc-item toc-level-2"><a class="toc-link" href="#%E5%A4%8D%E6%9D%82%E5%BA%A6%E5%88%86%E6%9E%90"><span class="toc-number">3.4.</span> <span class="toc-text"> 复杂度分析</span></a></li><li class="toc-item toc-level-2"><a class="toc-link" href="#%E5%8F%82%E8%80%83%E8%B5%84%E6%96%99"><span class="toc-number">3.5.</span> <span class="toc-text"> 参考资料</span></a></li></ol></li><li class="toc-item toc-level-1"><a class="toc-link" href="#leetcode-15-%E4%B8%89%E6%95%B0%E4%B9%8B%E5%92%8C"><span class="toc-number">4.</span> <span class="toc-text"> LeetCode 15. 三数之和</span></a><ol class="toc-child"><li class="toc-item toc-level-2"><a class="toc-link" href="#%E6%80%9D%E8%B7%AF-2"><span class="toc-number">4.1.</span> <span class="toc-text"> 思路</span></a></li><li class="toc-item toc-level-2"><a class="toc-link" href="#method-%E6%8E%92%E5%BA%8F-%E5%8F%8C%E6%8C%87%E9%92%88"><span class="toc-number">4.2.</span> <span class="toc-text"> Method: 排序 + 双指针</span></a></li></ol></li><li class="toc-item toc-level-1"><a class="toc-link" href="#leetcode-18-%E5%9B%9B%E6%95%B0%E4%B9%8B%E5%92%8C"><span class="toc-number">5.</span> <span class="toc-text"> LeetCode 18. 四数之和</span></a><ol class="toc-child"><li class="toc-item toc-level-2"><a class="toc-link" href="#%E6%80%9D%E8%B7%AF-3"><span class="toc-number">5.1.</span> <span class="toc-text"> 思路</span></a></li><li class="toc-item toc-level-2"><a class="toc-link" href="#method-%E6%8E%92%E5%BA%8F-%E5%8F%8C%E6%8C%87%E9%92%88-2"><span class="toc-number">5.2.</span> <span class="toc-text"> Method: 排序 + 双指针</span></a></li></ol></li><li class="toc-item toc-level-1"><a class="toc-link" href="#leetcode-202-%E5%BF%AB%E4%B9%90%E6%95%B0"><span class="toc-number">6.</span> <span class="toc-text"> LeetCode 202. 快乐数</span></a><ol class="toc-child"><li class="toc-item toc-level-2"><a class="toc-link" href="#%E6%80%9D%E8%B7%AF-4"><span class="toc-number">6.1.</span> <span class="toc-text"> 思路</span></a></li><li class="toc-item toc-level-2"><a class="toc-link" href="#method-1-%E5%93%88%E5%B8%8C-set"><span class="toc-number">6.2.</span> <span class="toc-text"> Method 1: 哈希 set</span></a></li><li class="toc-item toc-level-2"><a class="toc-link" href="#method-2-%E5%BF%AB%E6%85%A2%E6%8C%87%E9%92%88"><span class="toc-number">6.3.</span> <span class="toc-text"> Method 2: 快慢指针</span></a></li></ol></li><li class="toc-item toc-level-1"><a class="toc-link" href="#leetcode-242-%E6%9C%89%E6%95%88%E7%9A%84%E5%AD%97%E6%AF%8D%E5%BC%82%E4%BD%8D%E8%AF%8D"><span class="toc-number">7.</span> <span class="toc-text"> LeetCode 242. 有效的字母异位词</span></a><ol class="toc-child"><li class="toc-item toc-level-2"><a class="toc-link" href="#%E6%80%9D%E8%B7%AF-5"><span class="toc-number">7.1.</span> <span class="toc-text"> 思路</span></a></li></ol></li><li class="toc-item toc-level-1"><a class="toc-link" href="#leetcode-349-%E4%B8%A4%E4%B8%AA%E6%95%B0%E7%BB%84%E7%9A%84%E4%BA%A4%E9%9B%86"><span class="toc-number">8.</span> <span class="toc-text"> LeetCode 349. 两个数组的交集</span></a><ol class="toc-child"><li class="toc-item toc-level-2"><a class="toc-link" href="#%E6%80%9D%E8%B7%AF-6"><span class="toc-number">8.1.</span> <span class="toc-text"> 思路</span></a></li><li class="toc-item toc-level-2"><a class="toc-link" href="#unordered_set"><span class="toc-number">8.2.</span> <span class="toc-text"> unordered_set</span></a></li></ol></li><li class="toc-item toc-level-1"><a class="toc-link" href="#leetcode-383-%E8%B5%8E%E9%87%91%E4%BF%A1"><span class="toc-number">9.</span> <span class="toc-text"> LeetCode 383. 赎金信</span></a><ol class="toc-child"><li class="toc-item toc-level-2"><a class="toc-link" href="#method-%E5%93%88%E5%B8%8C"><span class="toc-number">9.1.</span> <span class="toc-text"> Method: 哈希</span></a></li></ol></li><li class="toc-item toc-level-1"><a class="toc-link" href="#leetcode-438-%E6%89%BE%E5%88%B0%E5%AD%97%E7%AC%A6%E4%B8%B2%E4%B8%AD%E6%89%80%E6%9C%89%E5%AD%97%E6%AF%8D%E5%BC%82%E4%BD%8D%E8%AF%8D"><span class="toc-number">10.</span> <span class="toc-text"> LeetCode 438. 找到字符串中所有字母异位词</span></a><ol class="toc-child"><li class="toc-item toc-level-2"><a class="toc-link" href="#method-%E6%BB%91%E5%8A%A8%E7%AA%97%E5%8F%A3%E6%B3%95"><span class="toc-number">10.1.</span> <span class="toc-text"> Method: 滑动窗口法</span></a></li></ol></li><li class="toc-item toc-level-1"><a class="toc-link" href="#leetcode-454-%E5%9B%9B%E6%95%B0%E7%9B%B8%E5%8A%A0ii"><span class="toc-number">11.</span> <span class="toc-text"> LeetCode 454. 四数相加 II</span></a><ol class="toc-child"><li class="toc-item toc-level-2"><a class="toc-link" href="#%E6%80%9D%E8%B7%AF-7"><span class="toc-number">11.1.</span> <span class="toc-text"> 思路</span></a></li><li class="toc-item toc-level-2"><a class="toc-link" href="#%E5%93%88%E5%B8%8C-map"><span class="toc-number">11.2.</span> <span class="toc-text"> 哈希 map</span></a></li></ol></li><li class="toc-item toc-level-1"><a class="toc-link" href="#leetcode-49-%E5%AD%97%E6%AF%8D%E5%BC%82%E4%BD%8D%E8%AF%8D%E5%88%86%E7%BB%84"><span class="toc-number">12.</span> <span class="toc-text"> LeetCode 49. 字母异位词分组</span></a><ol class="toc-child"><li class="toc-item toc-level-2"><a class="toc-link" href="#method-1-%E6%8E%92%E5%BA%8F-%E5%93%88%E5%B8%8C"><span class="toc-number">12.1.</span> <span class="toc-text"> Method 1: 排序 + 哈希</span></a></li><li class="toc-item toc-level-2"><a class="toc-link" href="#method-2-%E8%AE%A1%E6%95%B0-%E5%93%88%E5%B8%8C"><span class="toc-number">12.2.</span> <span class="toc-text"> Method 2: 计数 + 哈希</span></a></li></ol></li><li class="toc-item toc-level-1"><a class="toc-link" href="#leetcode-554-%E7%A0%96%E5%A2%99"><span class="toc-number">13.</span> <span class="toc-text"> LeetCode 554. 砖墙</span></a><ol class="toc-child"><li class="toc-item toc-level-2"><a class="toc-link" href="#method-%E5%93%88%E5%B8%8C-2"><span class="toc-number">13.1.</span> <span class="toc-text"> Method: 哈希</span></a></li><li class="toc-item toc-level-2"><a class="toc-link" href="#%E7%AE%97%E6%B3%95%E6%80%9D%E8%B7%AF-2"><span class="toc-number">13.2.</span> <span class="toc-text"> 算法思路</span></a></li><li class="toc-item toc-level-2"><a class="toc-link" href="#%E4%BB%A3%E7%A0%81%E5%AE%9E%E7%8E%B0-2"><span class="toc-number">13.3.</span> <span class="toc-text"> 代码实现</span></a></li><li class="toc-item toc-level-2"><a class="toc-link" href="#%E5%A4%8D%E6%9D%82%E5%BA%A6%E5%88%86%E6%9E%90-2"><span class="toc-number">13.4.</span> <span class="toc-text"> 复杂度分析</span></a></li><li class="toc-item toc-level-2"><a class="toc-link" href="#%E5%8F%82%E8%80%83%E8%B5%84%E6%96%99-2"><span class="toc-number">13.5.</span> <span class="toc-text"> 参考资料</span></a></li></ol></li></div><div class="related panel pjax" data-title="系列文章"><ul><li><a href="/leetcode-binarysearch.html" rel="bookmark" title="LeetCode - 二分查找专题">LeetCode - 二分查找专题</a></li><li><a href="/leetcode-vectors.html" rel="bookmark" title="LeetCode - 数组专题">LeetCode - 数组专题</a></li><li><a href="/leetcode-list.html" rel="bookmark" title="LeetCode - 链表专题">LeetCode - 链表专题</a></li><li class="active"><a href="/leetcode-hashtable.html" rel="bookmark" title="LeetCode - 哈希表专题">LeetCode - 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