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hyejj19
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문제 풀이 하시느라 고생많으셨습니다! 다음 한 주도 화이팅입니다~!!😃
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| # Approach: | ||
| # - Use a hash set to track elements we have already seen. | ||
| # - Iterate through the array and if the current number already exists in the set, return True. | ||
| # - Otherwise, add the current number to the set. |
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| for c in nums: | ||
| currentL = 1 | ||
| j = 1 | ||
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| while c+j in s: | ||
| s.remove(c+j) | ||
| currentL += 1 | ||
| j += 1 | ||
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| j = 1 | ||
| while c-j in s: | ||
| s.remove(c-j) | ||
| currentL += 1 | ||
| j += 1 | ||
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| longest = max(longest, currentL) |
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오! 기준 원소 c를 기준으로 양쪽을 제거해서 중복 탐색을 방지하는 의도군요!👍👍
근데 기준점인 c는 set에서 제거되지 않아서, 다음 반복에서 이웃 원소로 발견되어서 불필요한 탐색이 될 수도 있을 것 같습니다! for문에서 자기 자신을 제거하는 로직을 먼저 추가하고 시작하는건 어떨까요?
저는 이 문제를 c-1 이 없을 때 해당 원소를 연속된 숫자의 시작이라고 보고, 거기서부터 +1 하면서 longest 를 세는 방식으로 풀었는데 set을 제거 안하고 풀 수 있어서 좀 더 깔끔해질 수 있을 것 같습니다!
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제시해주신 방법이 더 나을것 같네요 감사합니다!
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