# 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

```python
class Solution:
    def findLadders(self, beginWord: str, endWord: str, wordList: List[str]) -> List[List[str]]:
        import collections
        ans = []
        steps = float("inf")
        if not beginWord or not endWord or not wordList or endWord not in wordList:
            return []
        word_dict = collections.defaultdict(list)
        L = len(beginWord)
        for word in wordList:
            for i in range(L):
                word_dict[word[:i] + "*" + word[i+1:]].append(word)
        queue = [[beginWord]]
        cost = {beginWord : 0}
        while queue:
            words_list= queue.pop(0)
            cur_word = words_list[-1]
            if cur_word == endWord:
                ans.append(words_list[:])
            else:
                for i in range(L):
                    intermediate_word = cur_word[:i] + "*" + cur_word[i+1:]
                    for word in word_dict[intermediate_word]:
                        w_l_temp = words_list[:]
                        if word not in cost or cost[word] >= cost[cur_word] + 1:
                            cost[word] = cost[cur_word] + 1
                            w_l_temp.append(word)
                            queue.append(w_l_temp[:])
                    
        return ans

```

## 答案

```python

```

## 选项

### A

```python

```

### B

```python

```

### C

```python

```