leetcode-problemset/leetcode-cn/originData/word-ladder.json

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        "question": {
            "questionId": "127",
            "questionFrontendId": "127",
            "categoryTitle": "Algorithms",
            "boundTopicId": 1442,
            "title": "Word Ladder",
            "titleSlug": "word-ladder",
            "content": "<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 -&gt; s<sub>1</sub> -&gt; s<sub>2</sub> -&gt; ... -&gt; s<sub>k</sub></code> such that:</p>\n\n<ul>\n\t<li>Every adjacent pair of words differs by a single letter.</li>\n\t<li>Every <code>s<sub>i</sub></code> for <code>1 &lt;= i &lt;= k</code> is in <code>wordList</code>. Note that <code>beginWord</code> does not need to be in <code>wordList</code>.</li>\n\t<li><code>s<sub>k</sub> == endWord</code></li>\n</ul>\n\n<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>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> beginWord = &quot;hit&quot;, endWord = &quot;cog&quot;, wordList = [&quot;hot&quot;,&quot;dot&quot;,&quot;dog&quot;,&quot;lot&quot;,&quot;log&quot;,&quot;cog&quot;]\n<strong>Output:</strong> 5\n<strong>Explanation:</strong> One shortest transformation sequence is &quot;hit&quot; -&gt; &quot;hot&quot; -&gt; &quot;dot&quot; -&gt; &quot;dog&quot; -&gt; cog&quot;, which is 5 words long.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> beginWord = &quot;hit&quot;, endWord = &quot;cog&quot;, wordList = [&quot;hot&quot;,&quot;dot&quot;,&quot;dog&quot;,&quot;lot&quot;,&quot;log&quot;]\n<strong>Output:</strong> 0\n<strong>Explanation:</strong> The endWord &quot;cog&quot; is not in wordList, therefore there is no valid transformation sequence.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= beginWord.length &lt;= 10</code></li>\n\t<li><code>endWord.length == beginWord.length</code></li>\n\t<li><code>1 &lt;= wordList.length &lt;= 5000</code></li>\n\t<li><code>wordList[i].length == beginWord.length</code></li>\n\t<li><code>beginWord</code>, <code>endWord</code>, and <code>wordList[i]</code> consist of lowercase English letters.</li>\n\t<li><code>beginWord != endWord</code></li>\n\t<li>All the words in <code>wordList</code> are <strong>unique</strong>.</li>\n</ul>\n",
            "translatedTitle": "单词接龙",
            "translatedContent": "<p>字典&nbsp;<code>wordList</code> 中从单词 <code>beginWord</code><em>&nbsp;</em>到&nbsp;<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\" />&nbsp;<code>beginWord -&gt; s<sub>1</sub>&nbsp;-&gt; s<sub>2</sub>&nbsp;-&gt; ... -&gt; s<sub>k</sub></code>：</p>\n\n<ul>\n\t<li>每一对相邻的单词只差一个字母。</li>\n\t<li><meta charset=\"UTF-8\" />&nbsp;对于&nbsp;<code>1 &lt;= i &lt;= k</code>&nbsp;时，每个<meta charset=\"UTF-8\" />&nbsp;<code>s<sub>i</sub></code>&nbsp;都在<meta charset=\"UTF-8\" />&nbsp;<code>wordList</code>&nbsp;中。注意， <code>beginWord</code><em>&nbsp;</em>不需要在<meta charset=\"UTF-8\" />&nbsp;<code>wordList</code>&nbsp;中。<meta charset=\"UTF-8\" /></li>\n\t<li><code>s<sub>k</sub>&nbsp;== endWord</code></li>\n</ul>\n\n<p>给你两个单词<em> </em><code>beginWord</code><em>&nbsp;</em>和 <code>endWord</code> 和一个字典 <code>wordList</code> ，返回 <em>从&nbsp;<code>beginWord</code> 到&nbsp;<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>\n&nbsp;\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<strong>输入：</strong>beginWord = \"hit\", endWord = \"cog\", wordList = [\"hot\",\"dot\",\"dog\",\"lot\",\"log\",\"cog\"]\n<strong>输出：</strong>5\n<strong>解释：</strong>一个最短转换序列是 \"hit\" -&gt; \"hot\" -&gt; \"dot\" -&gt; \"dog\" -&gt; \"cog\", 返回它的长度 5。\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<strong>输入：</strong>beginWord = \"hit\", endWord = \"cog\", wordList = [\"hot\",\"dot\",\"dog\",\"lot\",\"log\"]\n<strong>输出：</strong>0\n<strong>解释：</strong>endWord \"cog\" 不在字典中，所以无法进行转换。</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= beginWord.length &lt;= 10</code></li>\n\t<li><code>endWord.length == beginWord.length</code></li>\n\t<li><code>1 &lt;= wordList.length &lt;= 5000</code></li>\n\t<li><code>wordList[i].length == beginWord.length</code></li>\n\t<li><code>beginWord</code>、<code>endWord</code> 和 <code>wordList[i]</code> 由小写英文字母组成</li>\n\t<li><code>beginWord != endWord</code></li>\n\t<li><code>wordList</code> 中的所有字符串 <strong>互不相同</strong></li>\n</ul>\n",
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                    "code": "class Solution {\npublic:\n    int ladderLength(string beginWord, string endWord, vector<string>& wordList) {\n        \n    }\n};",
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                    "code": "class Solution(object):\n    def ladderLength(self, beginWord, endWord, wordList):\n        \"\"\"\n        :type beginWord: str\n        :type endWord: str\n        :type wordList: List[str]\n        :rtype: int\n        \"\"\"\n        ",
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