[LeetCode Sync] Runtime - 7 ms (52.69%), Memory - 19.4 MB (35.12%)

Author: NinePiece2

Solutions/0324-wiggle-sort-ii/README.md

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+<p>Given an integer array <code>nums</code>, reorder it such that <code>nums[0] &lt; nums[1] &gt; nums[2] &lt; nums[3]...</code>.</p>
+
+<p>You may assume the input array always has a valid answer.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums = [1,5,1,1,6,4]
+<strong>Output:</strong> [1,6,1,5,1,4]
+<strong>Explanation:</strong> [1,4,1,5,1,6] is also accepted.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums = [1,3,2,2,3,1]
+<strong>Output:</strong> [2,3,1,3,1,2]
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= nums.length &lt;= 5 * 10<sup>4</sup></code></li>
+	<li><code>0 &lt;= nums[i] &lt;= 5000</code></li>
+	<li>It is guaranteed that there will be an answer for the given input <code>nums</code>.</li>
+</ul>
+
+<p>&nbsp;</p>
+<strong>Follow Up:</strong> Can you do it in <code>O(n)</code> time and/or <strong>in-place</strong> with <code>O(1)</code> extra space?

Solutions/0324-wiggle-sort-ii/solution.py

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+class Solution:
+    def wiggleSort(self, nums: List[int]) -> None:
+        """
+        Do not return anything, modify nums in-place instead.
+        """
+        temp = sorted(nums)
+        n = len(temp)
+        i, j = (n - 1) >> 1, n - 1
+        for k in range(n):
+            if k % 2 == 0:
+                nums[k] = temp[i]
+                i -= 1
+            else:
+                nums[k] = temp[j]
+                j -= 1