[sora0349] Week 12 Solutions#1606
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sora0319 merged 5 commits intoDaleStudy:mainfrom Jun 22, 2025
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이번 한 주도 고생 많으셨습니다!! 다음 주도 화이팅이에요 🌻
| int end = intervals[i][1]; | ||
| if (prv_end > start) { | ||
| count++; | ||
| prv_end = Math.min(end, prv_end); |
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intervals를 start를 기준으로 오름차순 정렬한 후, 그리디하게 항상 Math.min()으로 더 작은 end를 선택하는 방식으로 풀이하신 것 같네요!
intervals를 end를 기준으로 오름차순 정렬하신다면 prv_end가 항상 end 보다 같거나 작을 것이기 때문에 Math.min() 로직을 제거할 수 있을 것 같습니다~!
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의견 주셔서 감사합니다
한번 더 생각해 봐야 할 것 같네요
| public class Solution { | ||
| public ListNode removeNthFromEnd(ListNode head, int n) { | ||
| Queue<ListNode> queue = new LinkedList<>(); | ||
| ListNode temp = new ListNode(0, head); |
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temp 노드를 이용함으로써 head가 삭제되는 케이스를 커버할 수 있는 좋은 풀이 방법인 것 같아요 🤩
다만, queue를 고정된 크기로 유지하며 O(n)의 공간을 사용하게 되는 부분을 투 포인터를 이용하여 슬라이딩 윈도우로 다룬다면 O(1)의 공간으로 최적화가 가능할 것 같습니다!
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공간 복잡도에 대해서는 생각하지 못했는데 투 포인터 방식으로 해볼 수 있겠네요!
| return false; | ||
| } | ||
| stack.push(new TreeNode[]{n1.left, n2.left}); | ||
| stack.push(new TreeNode[]{n1.right, n2.right}); |
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스택을 활용하면 이렇게 풀이할 수 있군요!! 저는 재귀로만 풀어보았는데, 참고가 되었습니다 😄
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