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[아현] WEEK02 Solutions #349
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f-exuan21
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Aug 19, 2024
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- Valid Anagram
- Counting Bits
- Encode and Decode Strings
- Construct Binary Tree From Preorder And Inorder Traversal
- Decode Ways
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charArray가 반복문을 위해서만 사용된 것 같아서
charArray를 생성하지 않고 chatAt을 이용하면 공간복잡도를 O(1)으로 푸는 것도 가능할 것 같은데 어떻게 생각하실까요?
추가적으로 n 이 무엇을 의미하는 걸까요?
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n 은 문자열의 길이를 나타냅니다.
말씀하신 데로 charAt을 쓰면 sToChar와 tToChar 를 사용하지 않아도 되어서 공간복잡도가 O(1)로 끝날 수 있어서 좋은 선택지인 것 같습니다! 👍
Counting bits solutions
주석 추가
공간복잡도를 O(n) -> O(1) 로 로직 변경
construct binary tree solutions
private int i = 0; | ||
Map<Integer, Integer> map = new HashMap<>(); |
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nit: 접근 제한자를 일관적으로 붙이거나 생략해주면 불필요한 오해의 소지를 없앨 수 있을 것 같아요.
private int i = 0; | |
Map<Integer, Integer> map = new HashMap<>(); | |
private int i = 0; | |
private Map<Integer, Integer> map = new HashMap<>(); |
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return true; | ||
} | ||
} |
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나머지 문제도 시간 되실 때 꼭 풀어보시면 좋을것 같습니다!! 배울게 많은 문제들이었던 것 같습니다=)
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3문제 푸시느라 고생 많으셨습니다! 마감 시간이 되었으니 PR 병합 부탁드리겠습니다.
@f-exuan21 님, 본 PR 계속 Draft 상태로 두시고 병합하지 않으시는 이유가 있으신가요? |
@DaleSeo 아 제가 주말동안 git에 접속을 못했습니다 ㅜㅠ 병합했습니다! |