127. Word Ladder（单词变换 广度优先）

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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.

For example,

Given:
beginWord = "hit"
endWord = "cog"
wordList = ["hot","dot","dog","lot","log","cog"]

As one shortest transformation is "hit" -> "hot" -> "dot" -> "dog" -> "cog",
return its length 5.

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.

UPDATE (2017/1/20):
The wordList parameter had been changed to a list of strings (instead of a set of strings). Please reload the code definition to get the latest changes.

这道题是经典的广度有优先搜索的例子，也是Dijkstra's algorithm的变形。
以题目给出的例子为例，其实就是在所有路径的权重都为1的情况下求出下列无向图中从节点hit到节点cog的最短路径：

 

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PS：图中相互之间只相差一个字母的单词都是相邻节点。

利用BFS算法，维持两个集合： visited 和 wordSet

 

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从hit开始出发，找到唯一的相邻节点：hot, 把它放进visited中,第一次循环结束。 PS： 所谓找到相邻的节点，在题目中就是找出和该单词之相差一个字母的所有单词。请看最后代码的详细实现。

 

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查看hot节点的所有相邻节点（已经被访问过的hit除外），找到lot和dot, 放进visited中。第二次循环结束。

 

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找出新家进来的lot和dot的未被访问的相邻节点，分别是log和dog放进visited中。第三次循环结束。

 

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找出log的未被访问的相邻节点cog，放进结合中。第四次循环结束。由于cog就是endWord，任务结束，跳出循环。

 

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这里总共经历了四次循环，每循环一次，其实就是从beginWord想endWord变化的一步，因此循环的次数（加上1）就是从beginWord想endWord转变经历的 number of steps。

class Solution:
    def ladderLength(self, beginWord, endWord, wordList):
        """
        :type beginWord: str
        :type endWord: str
        :type wordList: List[str]
        :rtype: int
        """
        wordList = set(wordList)
        visited = [beginWord]
        visited = set(visited)
        dist = 1
        
        while endWord not in visited:
            temp = set()
            for word in visited:
                for i in range(len(word)):
                    newwordL = list(word)
                    for ch in 'qwertyuiopasdfghjklzxcvbnm':
                        newwordL[i] = ch
                        newWord = ''.join(newwordL)
                        if newWord in wordList:
                            temp.add(newWord)
                            wordList.remove(newWord)

            dist += 1
            if len(temp) == 0: # if 0, it never gets to the endWord
                return 0

            visited = temp

        return dist

参考链接：https://www.jianshu.com/p/753bd585d57e