LeetCode-126、单词接龙 II-困难
给定两个单词（beginWord 和 endWord）和一个字典 wordList，找出所有从 beginWord 到 endWord 的最短转换序列。转换需遵循如下规则：
每次转换只能改变一个字母。
转换过程中的中间单词必须是字典中的单词。
说明：
如果不存在这样的转换序列，返回一个空列表。
所有单词具有相同的长度。
所有单词只由小写字母组成。
字典中不存在重复的单词。
你可以假设 beginWord 和 endWord 是非空的，且二者不相同。
示例 1：
输入:
beginWord = "hit",
endWord = "cog",
wordList = ["hot","dot","dog","lot","log","cog"]
输出:
[
["hit","hot","dot","dog","cog"],
["hit","hot","lot","log","cog"]
]示例 2：
输入:
beginWord = "hit"
endWord = "cog"
wordList = ["hot","dot","dog","lot","log"]
输出: []
解释: endWord "cog" 不在字典中，所以不存在符合要求的转换序列。代码：BFS + DFS
class Solution:
def findLadders(self, beginWord: str, endWord: str, wordList: List[str]) -> List[List[str]]:
word_bag = set(wordList)
if endWord not in word_bag:
return []
word_bag.add(beginWord)
distance = self.bfs(endWord, word_bag)
results = []
self.dfs(beginWord, endWord, word_bag, distance, [beginWord], results)
return results
def bfs(self, begin_word, word_bag):
queue = [begin_word]
distance = {}
distance[begin_word] = 0
while queue:
size = len(queue)
for _ in range(size):
curr_word = queue.pop(0)
for next_word in self.get_next_words(curr_word, word_bag):
if next_word not in distance:
distance[next_word] = distance[curr_word] + 1
queue.append(next_word)
return distance
def get_next_words(self, curr_word, word_bag):
next_words = []
for i in range(len(curr_word)):
for c in list(string.ascii_lowercase):
next_word = curr_word[:i] + c + curr_word[i+1:]
if next_word != curr_word and next_word in word_bag:
next_words.append(next_word)
return next_words
def dfs(self, curr_word, target, word_bag, distance, path, results):
if curr_word == target:
results.append(list(path))
return
for next_word in self.get_next_words(curr_word, word_bag):
if distance[next_word] != distance[curr_word] - 1:
continue
path.append(next_word)
self.dfs(next_word, target, word_bag, distance, path, results)
path.pop()
参考：
评论：BFS + DFS，代码如上
