[강희찬] WEEK 9 Solution #519
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TonyKim9401
reviewed
Oct 10, 2024
| function findMin(nums: number[]): number { | ||
| let left = 0; | ||
| let right = nums.length - 1; | ||
| let mid = (left + right) >> 1; |
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mid 값 구하는데 (left + right) / 2만 생각했는데, 이렇게 하면 비트 연산으로 더 효율적이겠네요..! 하나 배웠습니다 :)
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// 2 보다 우아하네요. mid에 대해 첨언드리자면 정수형 변수의 크기가 가변적이지 않은 언어의 경우 left + (right - left) // 2 같은 방식으로 mid를 계산해서 overflow를 방지하는 방식도 있습니다.
TonyKim9401
approved these changes
Oct 10, 2024
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| function hasCycle(head: ListNode | null): boolean { | ||
| const SET = new Set<ListNode>(); | ||
| while (head) { | ||
| if (SET.has(head)) return true; | ||
| SET.add(head); | ||
| head = head.next | ||
| } | ||
| return false; |
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fast와 slow만 사용하는걸 생각헀는데, 창의적인 답인것 같아요!
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| function maxSubArray(nums: number[]): number { | ||
| let maxSum = nums[0]; | ||
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| for (let i = 1; i < nums.length; i++) { | ||
| nums[i] = Math.max(nums[i], nums[i] + nums[i - 1]); | ||
| maxSum = Math.max(maxSum, nums[i]); | ||
| } | ||
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| return maxSum; |
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성능만 따지자면 세개의 답중 가장 느리고 공간 효율도 안좋지만, 가독성 측면에서 봤을때는 너무 좋은것 같아요. 개인적으로 이 코드가 제일 마음에 들어요 :)
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| * T.C. O(m * n) | ||
| * S.C. O(m * n) |
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시간 복잡도를 ^2이나 줄이시는 과정이 눈에 보여서 좋은것 같아요 :)
희찬의 경우에는 pacific, atlantic에서 공통 좌표를 가지고 있으면 추가해주는 식으로 하셨는데, 전 각각 boolean 타입의 배열을 만들어서, 둘다 true면 추가해주는 식으로 헀어요. 큰 차이는 없지만 제 풀이 코드도 확인해주시면 재미있으실것 같습니다 👍
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