127. Word Ladder Leetcode Python

Given two words (start and end), and a dictionary, find the length of shortest transformation sequence from start to end, such that:


Only one letter can be changed at a time
Each intermediate word must exist in the dictionary
For example,


Given:
start = "hit"
end = "cog"
dict = ["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.

这题是用的breadth first search 把每个字母当成一个edge 用一个que 把每次得到的新的edge加进来 在dict里面把已经遍历过的节点删除了。

每过一个新的单词的复杂度为L×26 L 是单词的长度

代码如下：

class Solution:    # @param start, a string    # @param end, a string    # @param dict, a set of string    # @return an integer    def ladderLength(self, start, end, dict):        dict.add(end)        q=[]        q.append((start,1))        while q:            curr=q.pop(0)            curword=curr[0]            curlen=curr[1]            if curword==end:                return curlen            for i in range(len(curword)):                part1=curword[:i]                part2=curword[i+1:]                for j in 'qwertyuiopasdfghjklzxcvbnm':                    if curword[i]!=j:                        nextword=part1+j+part2                        if nextword in dict:                            q.append((nextword,curlen+1))                            dict.remove(nextword)        return 0