[hellojoyworldz] WEEK 02 solutions#2697
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📊 hellojoyworldz 님의 학습 현황이번 주 제출 문제
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문제 풀이 현황
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parkhojeong
approved these changes
Jul 4, 2026
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| class Solution: | ||
| def isAnagram(self, s: str, t: str) -> bool: | ||
| return sorted(s) == sorted(t) | ||
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| for i in reversed(range(n)): | ||
| answer[i] *= right | ||
| right *= nums[i] |
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reversed(range(n)) 사용하신거 간결하고 좋네요~
| prev2, prev1 = 1, 1 | ||
| for _ in range(n - 1): | ||
| prev2, prev1 = prev1, prev2 + prev1 | ||
| return prev1 |
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개인적으로는 이런 부분들이 읽으면서 생각을 한 번 더 하게 되었던 거 같습니다!
- prev2, prev1에 1, 1 이 할당된 것도 문제의 조건과 어떻게 연결되는건지
- n = 1, n = 2 일때 어떻게 처리되는지
- prev1이 리턴하는게 n 값 까지 계산한 후 prev1 을 리턴하는 것인지
가독성을 고려한다면 배열을 사용해봐도 괜찮을 거 같습니다~
class Solution:
def climbStairs(self, n: int) -> int:
if n == 1:
return 1
arr = [0] * n
arr[0], arr[1] = 1, 2
for i in range(2, n):
arr[i] = arr[i - 1] + arr[i - 2]
return arr[-1]
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안 그래도 가독성 부분이 고민이였는데,
제안해 주신 것처럼 배열을 쓰니까 흐름이 훨씬 직관적이고 좋네요!
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| for i in range(len(nums)): | ||
| if i > 0 and nums[i] == nums[i - 1]: | ||
| continue |
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정렬되어 있는 상태라 nums[i] > 0 인 부분은 break해도 좋을 거 같습니다!
ex. [-1, 0, 1, 2, 3, 4, 5, 6, 7], i = 2 부터는 양수 + 양수 + 양수
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아 그러네요, 제가 놓쳤습니다
세심한 피드백 정말 감사합니다! 🙇♂️
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| if not (low < node.val < high): | ||
| return False |
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(log < node.val < high) 사용하신거 좋네요!
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