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hayashi-ay
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辞書を使っても良いので命名などはもう少しこだわると良いと思います。
| #対応表を作る | ||
| #.pop | ||
| dict_kakko = {')':'(','}':'{',']':'['} | ||
| open = ('(','{','[') |
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openはbuiltin functionなので変数名として使わないほうが良いでしょう。
https://docs.python.org/3/library/functions.html
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open_brackets, left_bracketsとかでしょうか。
| #openだったらstackに積む、closeだったら、stackの末尾のopenとcloseが対応しているかを確認。対応→〇、非対応→× | ||
| #対応表を作る | ||
| #.pop | ||
| dict_kakko = {')':'(','}':'{',']':'['} |
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変数名がないよりはマシですがローマ字読みの命名は好ましくないです。parentheses, bracketsあたりでしょうか。
| stack.append(kakko) | ||
| elif stack.pop() != dict_kakko[kakko]: | ||
| return False | ||
| break |
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01と同じですね
すみません、ありがとうございます
| elif stack.pop() != dict_kakko[kakko]: | ||
| return False | ||
| break | ||
| elif stack.pop() == dict_kakko[kakko]: |
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if-elif-elifでelseの場合がどうなるか読んでいて気になりました。実際にはこのelifはelif stack.pop() != dict_kakko[kakko]:の余事象になるので、elseで良いと思います。
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たしかに余事象になるので、わざわざelifで条件記載すると冗長になりますね
ありがとうございます
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https://leetcode.com/problems/valid-parentheses/