id： fuxuemingzhu

## 题目描述：

Given two words (beginWord and endWord), and a dictionary’s word list, find the length of shortest transformation sequence from beginWord to endWord, such that:

1. Only one letter can be changed at a time.
2. Each transformed word must exist in the word list. Note that beginWord is not a transformed word.

Note:

• Return 0 if there is no such transformation sequence.
• All words have the same length.
• All words contain only lowercase alphabetic characters.
• You may assume no duplicates in the word list.
• You may assume beginWord and endWord are non-empty and are not the same.

Example 1:

``````Input:
beginWord = "hit",
endWord = "cog",
wordList = ["hot","dot","dog","lot","log","cog"]

Output: 5

Explanation: As one shortest transformation is "hit" -> "hot" -> "dot" -> "dog" -> "cog",
return its length 5.
``````

Example 2:

``````Input:
beginWord = "hit"
endWord = "cog"
wordList = ["hot","dot","dog","lot","log"]

Output: 0
``````

Explanation: The endWord “cog” is not in wordList, therefore no possible transformation.

## 解题方法

``````class Solution(object):
"""
:type beginWord: str
:type endWord: str
:type wordList: List[str]
:rtype: int
"""
wordset = set(wordList)
if endWord not in wordset:
return 0
visited = set([beginWord])
chrs = [chr(ord('a') + i) for i in range(26)]
bfs = collections.deque([beginWord])
res = 1
while bfs:
len_bfs = len(bfs)
for _ in range(len_bfs):
origin = bfs.popleft()
for i in range(len(origin)):
originlist = list(origin)
for c in chrs:
originlist[i] = c
transword = "".join(originlist)
if transword not in visited:
if transword == endWord:
return res + 1
elif transword in wordset:
bfs.append(transword)
res += 1
return 0
``````

``````class Solution(object):
"""
:type beginWord: str
:type endWord: str
:type wordList: List[str]
:rtype: int
"""
wordset = set(wordList)
bfs = collections.deque()
bfs.append((beginWord, 1))
while bfs:
word, length = bfs.popleft()
if word == endWord:
return length
for i in range(len(word)):
for c in "abcdefghijklmnopqrstuvwxyz":
newWord = word[:i] + c + word[i + 1:]
if newWord in wordset and newWord != word:
wordset.remove(newWord)
bfs.append((newWord, length + 1))
return 0
``````

## 日期

2018 年 9 月 29 日 —— 国庆9天长假第一天！