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DaleSeo
reviewed
Oct 29, 2024
Comment on lines
+8
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| if (p == null && q == null) return true; | ||
| return false; |
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😉
Suggested change
| if (p == null && q == null) return true; | |
| return false; | |
| return p == null && q == null; |
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L7 ~ L10 을 조금 더 줄여서 아래처럼 가능할 것 같습니다!
if (p == null || q == null) {
return p == q;
}
SamTheKorean
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Nov 3, 2024
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전반적으로 코드가 읽기 편해서 알고리즘을 이해하기 편했습니다! 고생많으셨습니다!
bky373
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Nov 3, 2024
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정말 깔끔하게 풀어주셔서 코멘트 드릴 부분이 크게 없었네요!
한 주 동안 고생 많으셨습니다!
Comment on lines
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| if (currentEnd >= nextStart) { | ||
| currentInterval[1] = Math.max(currentEnd, nextEnd); |
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(사소) 이 조건이면 max 로 하지 않고, currentInterval[1] = currentEnd 로 해도 괜찮아 보입니다!
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아앗 nextStart 와 nextEnd 를 잘못 봤군요 위 내용은 넘어가시면 될 것 같습니다!
bky373
reviewed
Nov 3, 2024
| @@ -0,0 +1,17 @@ | |||
| // TC: O(n) | |||
| // retreive all given nodes | |||
| // SC: O(1) | |||
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dfs 스택을 사용하는 만큼 O(N) 이지 않을까요~?
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