使用字典，降低查找的復雜度。使用list會超時。

 1 class Solution:
 2 
 3     def nextWordsList(self, word, wordDict):
 4         res_list = []
 5         for i in range(len(word)):
 6             for j in string.ascii_lowercase:
 7                 new_word = list(word)
 8                 if j != word[i]:
 9                     new_word[i] = j
 
10                     new_word = ‘‘.join(new_word)
11                     if new_word in wordDict:
12                         res_list.append(new_word)
13                         del wordDict[new_word]
14         return res_list
15 
16 
17     def bfs(self, beginWord, endWord, wordDict):
18         #
 
 返回一個int
19         queue = []
20         queue.append([beginWord, 1])
21         while queue:
22             word, step = queue[0][0], queue[0][1]
23             queue.pop(0)
24             if word == endWord: return step
25             # 得到下一次變換一個單詞，得到的單詞列表
26             nextWords = self.nextWordsList(word, wordDict)
 
27             for j in nextWords:
28                 queue.append([j, step+1])
29         return 0
30     def ladderLength(self, beginWord, endWord, wordList):
31         """
32         :type beginWord: str
33         :type endWord: str
34         :type wordList: List[str]
35         :rtype: int
36         """
37         if beginWord in wordList: wordList.remove(beginWord)
38         wordDict = {}
39         for w in wordList: wordDict[w] = 1
40         return self.bfs(beginWord, endWord, wordDict)

LeetCode —— 單詞接龍（Python）