[leetcode127]Word Ladder Problem

Word Ladder Problem

问题描述：

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:

Only one letter can be changed at a time

Each intermediate word must exist in the word list

For example,

Given:

beginWord = “hit”

endWord = “cog”

wordList = [“hot”,”dot”,”dog”,”lot”,”log”]

As one shortest transformation is “hit” -> “hot” -> “dot” -> “dog” -> “cog”,

return its length 5.

Note:

Return 0 if there is no such transformation sequence.

All words have the same length.

All words contain only lowercase alphabetic characters.

算法：

搜索算法，求最短路径相关问题，应该用BFS。

只求路径长度，不需要具体路径。即要求遍历层数，用变量level标记层数，遇到终点返回层数level即可。

from collections import deque
class Solution(object):
def ladderLength(self, beginWord, endWord, wordDict):
"""
:type beginWord: str
:type endWord: str
:type wordDict: Set[str]
:rtype: int
"""
level = 1
que = deque()
visited = set()
wordDict.add(endWord)
que.append((beginWord, level)) # state is represented by (str, level depth)
while que:
curStr, curLevel = que.popleft()
curLevel += 1
for sucsr in self.successor(curStr, wordDict):
if sucsr == endWord:
return curLevel
if sucsr not in visited:
visited.add(sucsr)
wordDict.remove(sucsr)
que.append((sucsr,curLevel))
return 0

def successor(self, curStr, wordDict):
succss = []
curList = list(curStr)
for idx in range(len(curStr)):
for ele in 'abcdefghijklmnopqrstuvwxyz':
temp = curList[idx]
curList[idx] = ele
newStr = ''.join(curList)
curList[idx] = temp
if newStr in wordDict:
succss.append(newStr)
return succss

solution =  Solution()
wordSet = set(["hot","dot","dog","lot","log"])
start = 'hit'
end = 'cog'
print(solution.ladderLength(start,end,wordSet))

5