LeetCode —— 单词接龙（Python）

使用字典，降低查找的复杂度。使用list会超时。

 class Solution:

     def nextWordsList(self, word, wordDict):
         res_list = []
         for i in range(len(word)):
             for j in string.ascii_lowercase:
                 new_word = list(word)
                 if j != word[i]:
                     new_word[i] = j
                     new_word = ''.join(new_word)
                     if new_word in wordDict:
                         res_list.append(new_word)
                         del wordDict[new_word]
         return res_list

     def bfs(self, beginWord, endWord, wordDict):
         # 返回一个int
         queue = []
         queue.append([beginWord, 1])
         while queue:
             word, step = queue[0][0], queue[0][1]
             queue.pop(0)
             if word == endWord: return step
             # 得到下一次变换一个单词，得到的单词列表
             nextWords = self.nextWordsList(word, wordDict)
             for j in nextWords:
                 queue.append([j, step+1])
         return 0
     def ladderLength(self, beginWord, endWord, wordList):
         """
         :type beginWord: str
         :type endWord: str
         :type wordList: List[str]
         :rtype: int
         """
         if beginWord in wordList: wordList.remove(beginWord)
         wordDict = {}
         for w in wordList: wordDict[w] = 1
         return self.bfs(beginWord, endWord, wordDict)