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[재호] WEEK 02 Solutions #339
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// TC: O(N) - leetcode analyze 기준 | ||
// SC: O(N) | ||
There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. 혹시 배열 복재에 들어가는 메모리도 고려하셨을까요? There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. 평상시에 시간복잡도만 계산하고 공간복잡도는 대략적으로만 생각했는데 배열 복제의 메모리 사용도 고려해야하는지 몰랐습니다.. 😮 |
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/** | ||
* Definition for a binary tree node. | ||
* function TreeNode(val, left, right) { | ||
* this.val = (val===undefined ? 0 : val) | ||
* this.left = (left===undefined ? null : left) | ||
* this.right = (right===undefined ? null : right) | ||
* } | ||
*/ | ||
/** | ||
* @param {number[]} preorder | ||
* @param {number[]} inorder | ||
* @return {TreeNode} | ||
*/ | ||
var buildTree = function (preorder, inorder) { | ||
if (inorder.length === 0) { | ||
return null; | ||
} | ||
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const rootValue = preorder[0]; | ||
const leftNodeLength = inorder.findIndex((value) => value === rootValue); | ||
const leftNode = buildTree( | ||
preorder.slice(1, 1 + leftNodeLength), | ||
inorder.slice(0, leftNodeLength) | ||
); | ||
const rightNode = buildTree( | ||
preorder.slice(1 + leftNodeLength), | ||
inorder.slice(leftNodeLength + 1) | ||
); | ||
return new TreeNode(rootValue, leftNode, rightNode); | ||
}; |
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// TC: O(N) | ||
// SC: O(N) | ||
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/** | ||
* @param {number} n | ||
* @return {number[]} | ||
*/ | ||
var countBits = function (n) { | ||
const result = [0]; | ||
let pointer = 0; | ||
let lastPointer = 0; | ||
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for (let num = 1; num <= n; num++) { | ||
result.push(result[pointer] + 1); | ||
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if (pointer === lastPointer) { | ||
lastPointer = result.length - 1; | ||
pointer = 0; | ||
} else { | ||
pointer += 1; | ||
} | ||
} | ||
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return result; | ||
}; | ||
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There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. 이 투포인터 솔루션 신기한 것 같은데, 혹시 모임 때 소개해주실 수 있으실까요? There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. 달레님 알고리즘과 크게 다를 바 없다고 생각해서 조금 의아했는데 소개 이유를 들어보니 깨닫게 되었습니다! |
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// TC: O(N) | ||
// SC: O(N) | ||
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/** | ||
* @param {string} s | ||
* @return {number} | ||
*/ | ||
var numDecodings = function (s) { | ||
if (s[0] === "0") { | ||
return 0; | ||
} | ||
if (s.length === 1) { | ||
return 1; | ||
} | ||
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const dpTable = new Array(s.length).fill(0); | ||
if (s[0] !== "0") { | ||
dpTable[0] = 1; | ||
} | ||
if (s[1] !== "0") { | ||
dpTable[1] += 1; | ||
} | ||
if (isValid(`${s[0]}${s[1]}`)) { | ||
dpTable[1] += 1; | ||
} | ||
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for (let index = 2; index < s.length; index++) { | ||
if (s[index] !== "0") { | ||
dpTable[index] += dpTable[index - 1]; | ||
} | ||
if (s[index - 1] !== "0" && isValid(`${s[index - 1]}${s[index]}`)) { | ||
dpTable[index] += dpTable[index - 2]; | ||
} | ||
} | ||
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return dpTable[dpTable.length - 1]; | ||
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function isValid(stringNumber) { | ||
const number = Number(stringNumber); | ||
if (number <= 0) { | ||
return false; | ||
} | ||
if (27 <= number) { | ||
return false; | ||
} | ||
return true; | ||
} | ||
}; |
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// TC: O(N * log N) | ||
// SC: O(N) | ||
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/** | ||
* @param {string} s | ||
* @param {string} t | ||
* @return {boolean} | ||
*/ | ||
var isAnagram = function(s, t) { | ||
return s.split('').sort().join('') === t.split('').sort().join(''); | ||
There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. 오..... 나누고 정렬하고 합치고 좋네요 배워갑니다! |
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}; |
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복잡도 분석은 직접하시는 게 더 좋을 것 같습니다! 코딩 면접 때도 직접 하셔야하니까요 :)
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넵, leetcode analyze complexity에 의존하지 않고 직접 계산해보겠습니다!