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