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 possibletransformation.

Answer:

BFS,替換單詞中的字母

注意：使用set而不用list

class Solution(object):
    def ladderLength(self, beginWord, endWord, wordList):
        """
        :type beginWord: str
        :type endWord: str
        :type wordList: List[str]
        :rtype: int
        """
        wordListset=set(wordList)
        if endWord not in wordListset:
            return 0
        wl=len(beginWord)
        
        queue=[]
        queue.append((beginWord,1))
        visited={beginWord}
        
        while queue!=[]:
            #print queue
            (w,l)=queue.pop(0)
            
            for i in range(wl):
                for j in string.ascii_lowercase:
                    if j!=w[i]:
                        tempneww=w[0:i]+j+w[i+1:]
                        if tempneww in wordListset and tempneww not in visited:
                            if tempneww==endWord:
                                return l+1
                            queue.append((tempneww,l+1))
                            visited.add(tempneww)
                    
        return 0