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844. Backspace String Compare https://leetcode.com/problems/backspace-string-compare/
nodchip
reviewed
May 26, 2024
| ```python | ||
| class Solution: | ||
| def backspaceCompare(self, s: str, t: str) -> bool: | ||
| s_stack = [] |
| t_stack.pop() | ||
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| if len(s_stack) != len(t_stack): |
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list の == は内容まで比較しますので、 return s_stack == t_stack でよいと思います。
Owner
Author
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こちら知らなかったです。発見でした。
ありがとうございます。
| loop_cnt -= 1 | ||
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| return True | ||
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oda
reviewed
May 27, 2024
| t_stack = [] | ||
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| for char in s: | ||
| if char != '#': |
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char は言語によっては予約語なので、私は避けますが趣味の問題です。
同じ処理は二度書かずに関数にしましょう。
if 文字が '#' でないとき:
elif : # すなわち '#' なとき
と二重否定にしているのひねくれていませんか。
if char == '#':
if s_stack:
s_stack.pop()
continue
s_stack.append(char)
のほうが素直でしょう。
Owner
Author
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たしかに自分のコード、二重否定なのと無駄に否定系使ってるのが読みづらかったです。
頂いたコード読みやすかったです。
continueいまだに使い慣れてないので、論理の構造の幅のところを見直してみます。
oda
reviewed
May 27, 2024
| t_stack.pop() | ||
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| if len(s_stack) != len(t_stack): |
There was a problem hiding this comment.
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https://leetcode.com/problems/backspace-string-compare/