题目：原题链接（中等）
标签：字符串、广度优先遍历、回溯算法、图、图-无向图
| 解法 | 时间复杂度 | 空间复杂度 | 执行用时 |
|---|---|---|---|
| Ans 1 (Python) | O ( N × C ) O(N×C) O(N×C) | O ( N × C × 26 ) O(N×C×26) O(N×C×26) | 296ms (39.52%) |
| Ans 2 (Python) | |||
| Ans 3 (Python) |
解法一：
class Solution:
def ladderLength(self, beginWord: str, endWord: str, wordList: List[str]) -> int:
N = len(beginWord) # 单词长度
word_set = set(wordList) # 单词集合
# 处理目标词不存在于词典的情况
if endWord not in wordList:
return 0
# 初始化单词列表：将单词列表中每个单词的每个字符都替换为*，用以在O(C)的时间复杂度内计算邻边
# 当前步骤时间复杂度：N×C
word_hash = collections.defaultdict(list)
for word in wordList:
for i in range(N):
word_hash[word[:i] + "*" + word[i + 1:]].append(word)
# 寻找所有相邻结点
def near(word):
near_words = []
for ii in range(N): # 逐个字符遍历
for new_word in word_hash[word[:ii] + "*" + word[ii + 1:]]:
if new_word in word_set and new_word not in marked:
near_words.append(new_word)
return near_words
step = 1
marked = set() # 已访问的节点
queues = [[beginWord]] # 当前路径
ans = []
while queues:
new_queues = [] # 新的深度的路径
has_find_aim = False # 是否已找到目标词
# 将上一个深度的结点加入到已访问的节点
for queue in queues:
marked.add(queue[-1])
# 遍历寻找新的路径
for queue in queues:
for word in near(queue[-1]):
path = queue + [word]
if word == endWord:
ans.append(path)
has_find_aim = True
new_queues.append(path)
queues = new_queues
step += 1
if has_find_aim: # 判断是否已经找到目标词
return step
return 0
