[gitsunmin] Week 09 Solutions #527
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EgonD3V
approved these changes
Oct 12, 2024
| let right = nums.length - 1; | ||
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| while (left < right) { | ||
| const mid = Math.floor((left + right) / 2); |
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굳이 floor은 안사용하셔도 괜찮을 것 같습니다. 이분탐색에서 while문에 등호를 넣거나 if나 else에서 등호를 넣거나 테스트 해보시면 좋을 것 같습니다.
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| function hasCycle(head: ListNode | null): boolean { | ||
| if (!head || !head.next) { | ||
| return false; | ||
| } | ||
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| let slow: ListNode | null = head; | ||
| let fast: ListNode | null = head.next; | ||
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| while (slow !== fast) { | ||
| if (!fast || !fast.next) return false; | ||
| slow = slow!.next; | ||
| fast = fast.next.next; | ||
| } | ||
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| return true; |
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코드 리뷰를 하면서 드는 생각인데 union-find로도 풀어볼 걸 그랬네요
| const directions = [[1, 0], [-1, 0], [0, 1], [0, -1]]; | ||
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| function dfs(r: number, c: number, visited: boolean[][], prevHeight: number) { | ||
| if (r < 0 || c < 0 || r >= m || c >= n || visited[r][c] || heights[r][c] < prevHeight) { |
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저도 귀찮아서 한 줄로 쓰기는 하는데, 리뷰어의 입장에서 코드 스멜이나 단축 평가를 고려하면 적절한 조건들로 나누는 게 좋을 것 같습니다. 이 경우는 index error가 일어나지 않을 r, c를 한정하기 위해 순서가 어쩔 수 없지만...
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@lymchgmk
이야기 감사합니다! 코드 스멜도 신경써볼게요!
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