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liza0525
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Mar 19, 2026
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이번주도 고생하셨습니다! 이해하기 쉬운 방향으로 문제를 풀어주셔서 많이 배워갑니다!
| if ch.isalnum(): | ||
| cleaned_s += ch.lower() | ||
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| return cleaned_s == cleaned_s[::-1] |
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이런 풀이도 가능하겠네요! 코드가 깔끔해지는 풀이라고 생각되네요 :)
다만 문자열의 반까지만 나눠서 풀이를 하는 것도 가능하겠습니다~!
for i in range(len(cleaned_s // 2):
if cleaned_s[i] != cleaned_s [len(cleaned_s) -1 - i]:
return False # early return 처리!| return cleaned_s == cleaned_s[::-1] | ||
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| # Time Complexity : O(n) | ||
| # Space Complexity : O(n) |
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투포인터를 이용하면 공간복잡도도 O(1)까지 줄일 수 있습니다🙌
| @@ -0,0 +1,27 @@ | |||
| class Solution: | |||
| def hammingWeight(self, n: int) -> int: | |||
| # 1. Find highest power of 2 range that covers n | |||
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커버 가능한 2의 거듭제곱수를 먼저 구하고 역으로 계산하신 방법으로 푸신 거군요!
직관적으로 이해하기 쉬운 방법인 것 같습니다 👍🏼
10진수를 2진수로 변경하는 문제는 나누기 연산자와 모듈러 연산자로도 풀이가 가능해서
저는 그런 방식으로 풀어보았습니다!
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