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42 lines
2.0 KiB
HTML
42 lines
2.0 KiB
HTML
<p>Design a data structure that supports adding new words and finding if a string matches any previously added string.</p>
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<p>Implement the <code>WordDictionary</code> class:</p>
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<ul>
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<li><code>WordDictionary()</code> Initializes the object.</li>
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<li><code>void addWord(word)</code> Adds <code>word</code> to the data structure, it can be matched later.</li>
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<li><code>bool search(word)</code> Returns <code>true</code> if there is any string in the data structure that matches <code>word</code> or <code>false</code> otherwise. <code>word</code> may contain dots <code>'.'</code> where dots can be matched with any letter.</li>
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</ul>
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<p> </p>
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<p><strong>Example:</strong></p>
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<pre>
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<strong>Input</strong>
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["WordDictionary","addWord","addWord","addWord","search","search","search","search"]
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[[],["bad"],["dad"],["mad"],["pad"],["bad"],[".ad"],["b.."]]
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<strong>Output</strong>
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[null,null,null,null,false,true,true,true]
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<strong>Explanation</strong>
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WordDictionary wordDictionary = new WordDictionary();
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wordDictionary.addWord("bad");
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wordDictionary.addWord("dad");
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wordDictionary.addWord("mad");
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wordDictionary.search("pad"); // return False
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wordDictionary.search("bad"); // return True
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wordDictionary.search(".ad"); // return True
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wordDictionary.search("b.."); // return True
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</pre>
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<p> </p>
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<p><strong>Constraints:</strong></p>
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<ul>
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<li><code>1 <= word.length <= 25</code></li>
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<li><code>word</code> in <code>addWord</code> consists of lowercase English letters.</li>
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<li><code>word</code> in <code>search</code> consist of <code>'.'</code> or lowercase English letters.</li>
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<li>There will be at most <code>3</code> dots in <code>word</code> for <code>search</code> queries.</li>
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<li>At most <code>10<sup>4</sup></code> calls will be made to <code>addWord</code> and <code>search</code>.</li>
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</ul>
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