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oda
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Dec 26, 2025
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| static constexpr char kFail = 0; | ||
| static constexpr char kSuccess = 1; |
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個人的には、0, 1 で false, true を表すのは普通なのでこの定数はなくてもいいかなと感じます。
| can_segment.front() = kSuccess; | ||
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| for (int i = 1; i <= s_view.size(); ++i) { | ||
| for (auto word : word_dict) { |
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たしかに、そうですね。
辞書内の単語の長さは最大20なのでconst参照の方がよかったですね。
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| - https://github.com/katsukii/leetcode/pull/11/files | ||
| - Trieを使った解き方もある。 | ||
| - 実装が大変そうだったので、後でやろう。 |
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naoto-iwase
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Dec 26, 2025
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naoto-iwase
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選択肢を持って検討されていよいと思います。
すでにコメントが付いている点以外で読んでいて気になったことはありませんでした。
nodchip
reviewed
Dec 28, 2025
| class Solution { | ||
| public: | ||
| bool wordBreak(const string& s, const vector<string>& word_dict) { | ||
| vector<char> can_segment(s.size() + 1, kFalse); |
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少しお行儀が悪い感じがするのですが、
vector<char> can_segment(s.size() + 1, false);としても良いかなと思いました。
| if (i + word.size() > s.size()) { | ||
| continue; | ||
| } | ||
| if (s.substr(i, word.size()) != word) { |
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C++26 からは std::string::subview() 関数を使うことで、簡潔に書けるようです。
https://cpprefjp.github.io/reference/string/basic_string/subview.html
| if (s.substr(i, word.size()) != word) { | ||
| continue; | ||
| } | ||
| can_segment[i] = can_segment[i] || can_segment[i + word.size()]; |
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can_segment[i] |= can_segment[i + word.size()];のほうがシンプルだと思います。
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今回の問題
https://leetcode.com/problems/word-break/
次の問題
https://leetcode.com/problems/coin-change/