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[2023-08-14] dohyun #110 #122
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limstonestone
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Woo-Yeol,
ksumini and
zsmalla
as code owners
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Woo-Yeol
approved these changes
Aug 14, 2023
| stack = [] | ||
| tmp_ans = 1 # 곱하기 연산을 해야하므로 1부터 시작 | ||
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| def filter_correct(text): |
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오 예외처리를 정리하셔서 이렇게 filter로 걸어주시는게 너무 좋은 것 같습니다.
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혹시 (8) 이렇게 들어오는 케이스도 필터링이 되나요?
zsmalla
approved these changes
Aug 14, 2023
| stack = [] | ||
| tmp_ans = 1 # 곱하기 연산을 해야하므로 1부터 시작 | ||
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| def filter_correct(text): |
There was a problem hiding this comment.
스택의 유효성을 검증해주는 함수를 따로 정의하신 것 굉장히 좋은 것 같습니다.
다만 cond2에서 count는 O(N)이 소요되는 연산이므로, 모든 스택 연산이 끝났을 때 스택 리스트에 원소가 남아있다면 애초에 개수가 안맞았던 것 아니었을까? 로 접근해보면 어떨까 싶습니다!
ksumini
approved these changes
Aug 14, 2023
| tmp_ans = 1 # 곱하기 연산을 해야하므로 1부터 시작 | ||
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| def filter_correct(text): | ||
| cond1 = (text.count('(')==text.count(')')) & (text.count('[')==text.count(']')) |
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예외처리도 스터디 때 말씀 드렸던 것처럼 count의 시간복잡도는 O(n)
참고 링크: https://stackoverflow.com/questions/44812468/what-is-the-time-complexity-of-python-lists-count-function
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