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
