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KwonNayeon
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Feb 19, 2025
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#287 은 어떻게 풀어야할지 생각이 안나서 다음주 문제풀 때 해결해보겠습니다. |
KwonNayeon
approved these changes
Feb 21, 2025
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| /** | ||
| * Source: https://www.lintcode.com/problem/178/ | ||
| * Solution: 유효한 트리인지 순회하면서 확인하면 되기에 BFS로 구현 | ||
| * 시간 복잡도: O(V + E) - 노드와 간선에 한번은 방문 | ||
| * 공간 복잡도: O(V + E) - 인접리스트 만큼의 공간 필요 | ||
| */ |
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주석을 자세히 적어주셔서, 익숙하지 않은 언어인데도 알고리즘을 이해하기 수월했습니다!
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| function reorderList(head: ListNode | null): void { | ||
| if (!head || !head.next) return; | ||
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| // 1. 모든 노드를 배열에 저장 | ||
| const nodes: ListNode[] = []; | ||
| let current: ListNode | null = head; | ||
| while (current) { | ||
| nodes.push(current); | ||
| current = current.next; | ||
| } | ||
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| // 2. 배열의 양끝에서 시작하여 리스트 재구성 | ||
| let left = 0; | ||
| let right = nodes.length - 1; | ||
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| while (left < right) { | ||
| // 현재 왼쪽 노드의 다음을 저장 | ||
| nodes[left].next = nodes[right]; | ||
| left++; | ||
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| if (left === right) break; | ||
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| // 현재 오른쪽 노드를 다음 왼쪽 노드에 연결 | ||
| nodes[right].next = nodes[left]; | ||
| right--; | ||
| } | ||
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| // 마지막 노드의 next를 null로 설정 | ||
| nodes[left].next = null; | ||
| } |
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공간복잡도 개선을 위해 slow/fast 포인터를 사용한 방법도 도전해보시면 좋을 것 같습니다!
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@mike2ox 님 이번 주도 고생하셨습니다! 주석을 꼼꼼하게 적어주셔서 코드 리뷰가 수월했고, 공부도 많이 되었습니다. 덕분에 "Maximum depth of binary tree"를 반복문으로 접근해볼 수 있었습니다 😄
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