# [127. Word Ladder](https://leetcode.com/problems/word-ladder/)


## 题目

Given two words (*beginWord* and *endWord*), and a dictionary's word list, find the length of shortest transformation sequence from *beginWord* to *endWord*, such that:

1. Only one letter can be changed at a time.
2. Each transformed word must exist in the word list. Note that *beginWord* is *not* a transformed word.

**Note:**

- Return 0 if there is no such transformation sequence.
- All words have the same length.
- All words contain only lowercase alphabetic characters.
- You may assume no duplicates in the word list.
- You may assume *beginWord* and *endWord* are non-empty and are not the same.

**Example 1:**

    Input:
    beginWord = "hit",
    endWord = "cog",
    wordList = ["hot","dot","dog","lot","log","cog"]
    
    Output: 5
    
    Explanation: As one shortest transformation is "hit" -> "hot" -> "dot" -> "dog" -> "cog",
    return its length 5.

**Example 2:**

    Input:
    beginWord = "hit"
    endWord = "cog"
    wordList = ["hot","dot","dog","lot","log"]
    
    Output: 0
    
    Explanation: The endWord "cog" is not in wordList, therefore no possible transformation.


## 题目大意

给定两个单词（beginWord 和 endWord）和一个字典，找到从 beginWord 到 endWord 的最短转换序列的长度。转换需遵循如下规则：

1. 每次转换只能改变一个字母。
2. 转换过程中的中间单词必须是字典中的单词。

说明:

- 如果不存在这样的转换序列，返回 0。
- 所有单词具有相同的长度。
- 所有单词只由小写字母组成。
- 字典中不存在重复的单词。
- 你可以假设 beginWord 和 endWord 是非空的，且二者不相同。


## 解题思路

- 这一题要求输出从 `beginWord` 变换到 `endWord` 最短变换次数。可以用 BFS，从 `beginWord` 开始变换，把该单词的每个字母都用 `'a'~'z'` 变换一次，生成的数组到 `wordList` 中查找，这里用 Map 来记录查找。找得到就入队列，找不到就输出 0 。入队以后按照 BFS 的算法依次遍历完，当所有单词都 `len(queue)<=0` 出队以后，整个程序结束。
- 这一题题目中虽然说了要求找到一条最短的路径，但是实际上最短的路径的寻找方法已经告诉你了：
	1. 每次只变换一个字母
	2. 每次变换都必须在 `wordList` 中  
所以不需要单独考虑何种方式是最短的。