120. 单词接龙 (BFS)

描述

给出两个单词（start和end）和一个字典，找到从start到end的最短转换序列

比如：

每次只能改变一个字母。

变换过程中的中间单词必须在字典中出现。

如果没有转换序列则返回0。

所有单词具有相同的长度。

所有单词都只包含小写字母。

样例

给出数据如下：

start = "hit"

end = "cog"

dict = ["hot","dot","dog","lot","log"]

一个最短的变换序列是 "hit" -> "hot" -> "dot" -> "dog" -> "cog"，

返回它的长度 5

import collections

class Solution:
    # @param start, a string
    # @param end, a string
    # @param dict, a set of string
    # @return an integer
    def ladderLength(self, start, end, dict):
        # write your code here
        dict.add(end)
        wordLen = len(start)
        queue = collections.deque([(start, 1)])
        while queue:
            curr = queue.popleft()
            currWord = curr[0]; currLen = curr[1]
            if currWord == end: return currLen
            for i in range(wordLen):
                part1 = currWord[:i]; part2 = currWord[i+1:]
                for j in 'abcdefghijklmnopqrstuvwxyz':
                    if currWord[i] != j:
                        nextWord = part1 + j + part2
                        if nextWord in dict:
                            queue.append((nextWord, currLen + 1))
                            dict.remove(nextWord)
        return 0