139. Word Break#32
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liquo-rice
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May 14, 2026
| @@ -0,0 +1,22 @@ | |||
| # Step 2 | |||
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| - DPテーブルの変数、良い名付け方が思いつかない | |||
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眺める程度しかできていなかったです。
該当dpの変数名に関しては、他の方のを見る時意識していたのですが、しっくり来なかったです。
単純にdpと命名するのが良かったのかもしれません。
| ```python | ||
| class Solution: | ||
| def wordBreak(self, s: str, wordDict: List[str]) -> bool: | ||
| boolean_list = [False] * (len(s) + 1) |
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| for i in range(len(s) - 1, -1, -1): | ||
| for word in wordDict: | ||
| if i + len(word) <= len(s) and s[i:i + len(word)] in wordDict: |
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s[i : i + len(word)] == wordで良いと思います。NeetCodeの動画でもそうなっていました。
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これはおっしゃる通りです。in wordDictだとwordDictのサイズ分計算量が増えてしまいます。
リファクタの際に見逃していました。
| for word in wordDict: | ||
| if i + len(word) <= len(s) and s[i:i + len(word)] in wordDict: | ||
| boolean_list[i] = boolean_list[i + len(word)] | ||
| if boolean_list[i]: |
| return boolean_list[0] | ||
| ``` | ||
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| 時間計算量: $O(nmt)$ |
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変数の定義がないですね。ただ、s[i:i + len(word)] in wordDictを見ると違うように見えます。
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s[i:i + len(word)] == wordで実装できていればO(nmt)になっていましたが、私が行なった実装だとO(nm^2t)になってしまうと思います。
n: sのサイズ
m: wordDictのサイズ
t: wordDict内の単語の最大サイズ
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