[gitsunmin] Week15 Solutions #610
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HC-kang
approved these changes
Nov 23, 2024
| /** | ||
| * https://leetcode.com/problems/longest-palindromic-substring/ | ||
| * time complexity : O(n) | ||
| * space complexity : O(n^2) |
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시간, 공간 복잡도를 이렇게 판단하신 이유가 있으실까요?
제 생각에 시간 복잡도는 최대 N * N/2로 O(n^2) 이고 공간복잡도는 리턴값을 고려하더라도 O(n), 제외한다면 O(1)로 볼 수 있을 것 같아서요!
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이렇게 계산했어요!
• for 루프: 문자열 길이 n에 따라 n번 반복 → O(n).
• 팰린드롬 확장: 각 중심에서 최대 문자열 길이만큼 확장 → O(n).
• 전체 시간 복잡도: O(n) × O(n) = O(n²).
최악의 경우, 모든 중심에서 최대 확장되므로 O(n²)
| function longestPalindrome(s: string): string { | ||
| if (s.length <= 1) return s; | ||
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| let longestPalindromeStartIndex = 0; |
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중요한 사항은 아닙니다!
변수명이 조금 지나친 것 같습니다.. 읽기 어려워요ㅠㅠ
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| function isValidBST(root: TreeNode | null): boolean { | ||
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| return validate(root, -Infinity, Infinity); |
EgonD3V
reviewed
Nov 23, 2024
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| function isSameTree(p: TreeNode | null, q: TreeNode | null): boolean { | ||
| if (!p && !q) return true; | ||
| if (!p || !q || p.val !== q.val) return false; |
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재귀함수의 early return에서 조심할 점이 있다면 단축평가에서 손해보는 점인 것 같습니다. 그리고 if-else가 아니라 if문 단독으로 쓰는 경우, 개인적으로 각 if 문들의 조건이 무조건 독립적이도록 짜면 어떨까 싶습니다.
EgonD3V
approved these changes
Nov 23, 2024
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