392. Is Subsequence #57
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nodchip
reviewed
Aug 26, 2025
| ++it; | ||
| } | ||
| if (it == s.end()) { | ||
| break; |
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ここで return true、最後で return false してもよいと思います。
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実は最初そのように書いたのですが、leetcode上でs = "", t = ""のテストケースがあり、それに対応するためにこのようにしていました。
tokuhirat
reviewed
Aug 26, 2025
| }; | ||
| ``` | ||
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| イテレーターで書くよりintとかsize_tでインデックスを保存してoperator[]で要素比較したほうが読みやすいですかね？ |
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良いと思いました。
イテレーターで書くメリットは、string& 以外の型に拡張する時に楽、添字で配列外参照しない、ぐらいでしょうか。
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確かに今回でいうs, tなどの変数名に応じてプログラムを変えなくていいのはメリットですね！
tokuhirat
reviewed
Aug 26, 2025
| Follow upあるの気づかなかった。 | ||
| Follow up: Suppose there are lots of incoming s, say s1, s2, ..., sk where k >= 109, and you want to check one by one to see if t has its subsequence. In this scenario, how would you change your code? | ||
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| tにおいてある文字が出てくるindexの配列を<char, vector<int>>の辞書として管理しておいて、現在の場所以降で一番最初にあるs中の文字が出てくる場所を二分探索で探す方法（1つのsあたりの時間計算量O(|s|log|s|)）の他に、 |
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読みやすいと思います。 |
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https://leetcode.com/problems/is-subsequence/description/