Nav: << previous: 126.单词接龙 II | next: 128.最长连续序列 >>
Description
tab: English
<p>A <strong>transformation sequence</strong> from word <code>beginWord</code> to word <code>endWord</code> using a dictionary <code>wordList</code> is a sequence of words <code>beginWord -> s<sub>1</sub> -> s<sub>2</sub> -> ... -> s<sub>k</sub></code> such that:</p>
<ul>
<li>Every adjacent pair of words differs by a single letter.</li>
<li>Every <code>s<sub>i</sub></code> for <code>1 <= i <= k</code> is in <code>wordList</code>. Note that <code>beginWord</code> does not need to be in <code>wordList</code>.</li>
<li><code>s<sub>k</sub> == endWord</code></li>
</ul>
<p>Given two words, <code>beginWord</code> and <code>endWord</code>, and a dictionary <code>wordList</code>, return <em>the <strong>number of words</strong> in the <strong>shortest transformation sequence</strong> from</em> <code>beginWord</code> <em>to</em> <code>endWord</code><em>, or </em><code>0</code><em> if no such sequence exists.</em></p>
<p> </p>
<p><strong class="example">Example 1:</strong></p>
<pre>
<strong>Input:</strong> beginWord = "hit", endWord = "cog", wordList = ["hot","dot","dog","lot","log","cog"]
<strong>Output:</strong> 5
<strong>Explanation:</strong> One shortest transformation sequence is "hit" -> "hot" -> "dot" -> "dog" -> cog", which is 5 words long.
</pre>
<p><strong class="example">Example 2:</strong></p>
<pre>
<strong>Input:</strong> beginWord = "hit", endWord = "cog", wordList = ["hot","dot","dog","lot","log"]
<strong>Output:</strong> 0
<strong>Explanation:</strong> The endWord "cog" is not in wordList, therefore there is no valid transformation sequence.
</pre>
<p> </p>
<p><strong>Constraints:</strong></p>
<ul>
<li><code>1 <= beginWord.length <= 10</code></li>
<li><code>endWord.length == beginWord.length</code></li>
<li><code>1 <= wordList.length <= 5000</code></li>
<li><code>wordList[i].length == beginWord.length</code></li>
<li><code>beginWord</code>, <code>endWord</code>, and <code>wordList[i]</code> consist of lowercase English letters.</li>
<li><code>beginWord != endWord</code></li>
<li>All the words in <code>wordList</code> are <strong>unique</strong>.</li>
</ul>
---
[submissions](https://leetcode.com/problems/word-ladder/submissions/) | [solutions](https://leetcode.com/problems/word-ladder/solutions/)
tab: 中文
<p>字典 <code>wordList</code> 中从单词 <code>beginWord</code><em> </em>到 <code>endWord</code> 的 <strong>转换序列 </strong>是一个按下述规格形成的序列<meta charset="UTF-8" /> <code>beginWord -> s<sub>1</sub> -> s<sub>2</sub> -> ... -> s<sub>k</sub></code>：</p>
<ul>
<li>每一对相邻的单词只差一个字母。</li>
<li><meta charset="UTF-8" /> 对于 <code>1 <= i <= k</code> 时，每个<meta charset="UTF-8" /> <code>s<sub>i</sub></code> 都在<meta charset="UTF-8" /> <code>wordList</code> 中。注意， <code>beginWord</code><em> </em>不需要在<meta charset="UTF-8" /> <code>wordList</code> 中。<meta charset="UTF-8" /></li>
<li><code>s<sub>k</sub> == endWord</code></li>
</ul>
<p>给你两个单词<em> </em><code>beginWord</code><em> </em>和 <code>endWord</code> 和一个字典 <code>wordList</code> ，返回 <em>从 <code>beginWord</code> 到 <code>endWord</code> 的 <strong>最短转换序列</strong> 中的 <strong>单词数目</strong></em> 。如果不存在这样的转换序列，返回 <code>0</code> 。</p>
<p><strong>示例 1：</strong></p>
<pre>
<strong>输入：</strong>beginWord = "hit", endWord = "cog", wordList = ["hot","dot","dog","lot","log","cog"]
<strong>输出：</strong>5
<strong>解释：</strong>一个最短转换序列是 "hit" -> "hot" -> "dot" -> "dog" -> "cog", 返回它的长度 5。
</pre>
<p><strong>示例 2：</strong></p>
<pre>
<strong>输入：</strong>beginWord = "hit", endWord = "cog", wordList = ["hot","dot","dog","lot","log"]
<strong>输出：</strong>0
<strong>解释：</strong>endWord "cog" 不在字典中，所以无法进行转换。</pre>
<p> </p>
<p><strong>提示：</strong></p>
<ul>
<li><code>1 <= beginWord.length <= 10</code></li>
<li><code>endWord.length == beginWord.length</code></li>
<li><code>1 <= wordList.length <= 5000</code></li>
<li><code>wordList[i].length == beginWord.length</code></li>
<li><code>beginWord</code>、<code>endWord</code> 和 <code>wordList[i]</code> 由小写英文字母组成</li>
<li><code>beginWord != endWord</code></li>
<li><code>wordList</code> 中的所有字符串 <strong>互不相同</strong></li>
</ul>
---
[提交记录](https://leetcode.cn/problems/word-ladder/submissions/) | [题解](https://leetcode.cn/problems/word-ladder/solution/)
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