[thispath98] Week 4 #821
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[thispath98] Week 4 #821
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mmyeon
approved these changes
Jan 3, 2025
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@thispath98 님, 안녕하세요~
이번 주도 문제 푸시느라 정말 고생많으셨습니다.
코드 의도랑 복잡도 모두 자세히 남겨주셔서 리뷰하면서 큰 도움이 되었습니다!
새해 복 많이 많으시고, 다음 주도 함께 파이팅해봐요!
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| while True: | ||
| if list1 is None: | ||
| current_node.next = list2 | ||
| break | ||
| elif list2 is None: | ||
| current_node.next = list1 | ||
| break |
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while True로 주셨는데, 위의 코드에 남겨주신 것처럼 조건 명시하면 더 좋을 것 같아요 ~
while list1 is not None and list2 is not None:
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감사합니다. 무한 루프는 최소화해보겠습니다!
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| num_set = set(nums) | ||
| for i in range(len(nums) + 1): | ||
| if i not in num_set: | ||
| return i |
|
@mmyeon 님 리뷰 감사드립니다! |
mmyeon
reviewed
Jan 7, 2025
| O(amount): | ||
| amount만큼의 크기를 가지는 dp 배열을 저장한다. | ||
| """ | ||
| dp = [0 for _ in range(amount + 1)] |
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dp의 요소 초기값을 0이 아니라 최대값으로 주면 어떨까요?
저도 이번에 배웠는데, 최소 개수 구하는 문제라서 최대값으로 초기화하니까 if문도 줄고, 로직이 더 명확해지더라구요!!
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