题目地址: https://leetcode.com/problems/word-ladder/description/
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、 Onlyonelettercanbechangedatatime.；
2、 Eachtransformedwordmustexistinthewordlist.NotethatbeginWordisnotatransformedword.；
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.
题目大意
这个题名字是词语梯子，简单理解就是从begin开始，每次只能替换已经转化了的单词的其中一个字符，看最终能不能得到end。有个要求就是，每次变化不是任意的，是必须变成wordList中的其中一个才行。
解题方法
拿到这个题没有什么思路，看了别人解答之后，才猛然发现这个题是走迷宫问题的变形！也就是说，我们每次变化有26个方向，如果变化之后的位置在wordList中，我们认为这个走法是合规的，最后问能不能走到endWord？
很显然这个问题是BFS的问题，只是把走迷宫问题的4个方向转变成了26个方向，直接BFS会超时，所以我使用了个visited来保存已经遍历了的字符串，代表已经走过了的位置。代码总体思路很简单，就是利用队列保存每个遍历的有效的字符串，然后对队列中的每个字符串再次遍历，保存每次遍历的长度即可。
时间复杂度是O(NL)，空间复杂度是O(N).其中N是wordList中的单词个数，L是其实字符串的长度。
class Solution(object):
def ladderLength(self, beginWord, endWord, wordList):
"""
: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)
visited.add(transword)
res += 1
return 0
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显然上面的这个做法还是可以变短一点的，想起之前的二叉树的BFS的时候，会在每个节点入队列的时候同时保存了这个节点的深度，这样就少了一层对bfs当前长度的循环，可以使得代码变短。同时，学会了一个技巧，直接把已经遍历过的位置从wordList中删除，这样就相当于我上面的那个visited数组。下面这个代码很经典了，可以记住。
class Solution(object):
def ladderLength(self, beginWord, endWord, wordList):
"""
: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
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参考资料：
http://www.cnblogs.com/grandyang/p/4539768.html
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