# 单词接龙 II

<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 <= 7</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> 中的所有单词 <strong>互不相同</strong></li>
</ul>
</div>
</div>


## template

```cpp
#include <bits/stdc++.h>
using namespace std;

class Solution
{
public:
    bool differ_one(string &s, string &t)
    {
        int n = 0;
        for (int i = 0; i < s.size(); ++i)
            if ((n += s[i] != t[i]) > 1)
                return false;
        return n == 1;
    }
    vector<vector<int>> fa;
    vector<string> tmp;
    vector<vector<string>> ans;
    void dfs(int index, vector<string> &wordList)
    {
        if (fa[index].empty())
        {
            reverse(tmp.begin(), tmp.end());
            ans.push_back(tmp);
            reverse(tmp.begin(), tmp.end());
        }
        for (auto &x : fa[index])
        {
            tmp.push_back(wordList[x]);
            dfs(x, wordList);
            tmp.pop_back();
        }
    }
    vector<vector<string>> findLadders(string beginWord, string endWord, vector<string> &wordList)
    {

        int b = -1, e = -1, x;
        for (int i = 0; i < wordList.size(); ++i)
        {
            if (wordList[i] == beginWord)
                b = i;
            if (wordList[i] == endWord)
                e = i;
        }
        if (e == -1)
            return ans;
        if (b == -1)
            wordList.push_back(beginWord), b = wordList.size() - 1;
        queue<int> que;
        fa.assign(wordList.size(), vector<int>());
        vector<int> index(wordList.size(), 0);
        que.push(b), index[b] = 1;
        while (!que.empty())
        {
            x = que.front(), que.pop();
            if (index[e] && index[x] >= index[e])
                break;
            for (int i = 0; i < wordList.size(); ++i)
                if ((index[i] == 0 || index[x] + 1 == index[i]) && differ_one(wordList[x], wordList[i]))
                    if (index[i] == 0)
                        index[i] = index[x] + 1, que.push(i), fa[i].push_back(x);
                    else
                        fa[i].push_back(x);
        }
        if (index[e] == 0)
            return ans;
        tmp.push_back(endWord);
        dfs(e, wordList);
        tmp.pop_back();
        return ans;
    }
};

```

## 答案

```cpp

```

## 选项

### A

```cpp

```

### B

```cpp

```

### C

```cpp

```