Ｗord Ladder

这题主要是换个思路会简单很多。不用ＢＦＳ而用ＤＦＳ

class Solution(object):
    def ladderLength(self, beginWord, endWord, wordList):
        wordList.add(endWord)
        queue = collections.deque([[beginWord, 1]])
        while queue:
            word, length = queue.popleft()
            if word == endWord:
                return length
            for i in range(len(word)):
                for c in 'abcdefghijklmnopqrstuvwxyz':
                    next_word = word[:i] + c + word[i+1:]
                    if next_word in wordList:
                        wordList.remove(next_word)
                        queue.append([next_word, length + 1])
        return 0