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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>all the <strong>shortest transformation sequences</strong> from</em> <code>beginWord</code> <em>to</em> <code>endWord</code><em>, or an empty list if no such sequence exists. Each sequence should be returned as a list of the words </em><code>[beginWord, s<sub>1</sub>, s<sub>2</sub>, ..., s<sub>k</sub>]</code>.</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> [["hit","hot","dot","dog","cog"],["hit","hot","lot","log","cog"]]
<strong>Explanation:</strong> There are 2 shortest transformation sequences:
"hit" -> "hot" -> "dot" -> "dog" -> "cog"
"hit" -> "hot" -> "lot" -> "log" -> "cog"
</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> []
<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 <= 5</code></li>
<li><code>endWord.length == beginWord.length</code></li>
<li><code>1 <= wordList.length <= 500</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>
<li>The <strong>sum</strong> of all shortest transformation sequences does not exceed <code>10<sup>5</sup></code>.</li>
</ul>
---
[submissions](https://leetcode.com/problems/word-ladder-ii/submissions/) | [solutions](https://leetcode.com/problems/word-ladder-ii/solutions/)
tab: 中文
<p>按字典 <code>wordList</code> 完成从单词 <code>beginWord</code> 到单词 <code>endWord</code> 转化，一个表示此过程的 <strong>转换序列</strong> 是形式上像 <code>beginWord -> s<sub>1</sub> -> s<sub>2</sub> -> ... -> s<sub>k</sub></code> 这样的单词序列，并满足：</p>
<div class="original__bRMd">
<div>
<ul>
<li>每对相邻的单词之间仅有单个字母不同。</li>
<li>转换过程中的每个单词 <code>s<sub>i</sub></code>（<code>1 <= i <= k</code>）必须是字典 <code>wordList</code> 中的单词。注意，<code>beginWord</code> 不必是字典 <code>wordList</code> 中的单词。</li>
<li><code>s<sub>k</sub> == endWord</code></li>
</ul>
<p>给你两个单词 <code>beginWord</code> 和 <code>endWord</code> ，以及一个字典 <code>wordList</code> 。请你找出并返回所有从 <code>beginWord</code> 到 <code>endWord</code> 的 <strong>最短转换序列</strong> ，如果不存在这样的转换序列，返回一个空列表。每个序列都应该以单词列表<em> </em><code>[beginWord, s<sub>1</sub>, s<sub>2</sub>, ..., s<sub>k</sub>]</code> 的形式返回。</p>
<p> </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>[["hit","hot","dot","dog","cog"],["hit","hot","lot","log","cog"]]
<strong>解释：</strong>存在 2 种最短的转换序列：
"hit" -> "hot" -> "dot" -> "dog" -> "cog"
"hit" -> "hot" -> "lot" -> "log" -> "cog"
</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>[]
<strong>解释：</strong>endWord "cog" 不在字典 wordList 中，所以不存在符合要求的转换序列。
</pre>
<p> </p>
<p><strong>提示：</strong></p>
<ul>
<li><code>1 <= beginWord.length <= 5</code></li>
<li><code>endWord.length == beginWord.length</code></li>
<li><code>1 <= wordList.length <= 500</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> 中的所有单词 <strong>互不相同</strong></li>
</ul>
</div>
</div>
---
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