392. is subsequence#3
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lightbanana
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Jun 16, 2026
lightbanana
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- 今回解いた問題:392. Is Subsequence https://leetcode.com/problems/is-subsequence?envType=problem-list-v2&envId=xo2bgr0r
- 次に解く問題:141. Linked List Cycle https://leetcode.com/problems/linked-list-cycle?envType=problem-list-v2&envId=xo2bgr0r
nodchip
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Jun 18, 2026
| class Solution: | ||
| def isSubsequence(self, s: str, t: str) -> bool: | ||
| index = 0 | ||
| if len(s) == 0: |
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こちらのコメントをご参照ください。
mamo3gr/arai60#6 (comment)
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ありがとうございます！ Implicit False を用いることでコードが簡潔になることを確認しました。これからは使おうと思います。
nodchip
reviewed
Jun 18, 2026
| def isSubsequence(self, s: str, t: str) -> bool: | ||
| pattern = "" | ||
| for c in s: | ||
| pattern += ".*" + c |
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この書き方の場合、時間計算量が O(N2) かかる点に注意が必要です。 Python は文字列は Immutable で、文字列を連結するたびに新しいインスタンスが作られます。新しいインスタンスが作られる度に、既存の文字列をコピーし、新しい文字を連結するため、トータルで O(N2) かかります。
特定の条件下で in-place な処理に置き換わりますが、置き換わる条件は常識には含まれていないと思います。これを前提とした処理を書くのは避けたほうが無難だと思います。
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ありがとうございます。Pythonの文字列の連結は新しいインスタンスが作られるたびに既存の文字列をコピーし、時間計算量がO(n^2)になるため、避けることが無難であることを理解しました。
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