leetcode @python 127. Word Ladder

题目链接

https://leetcode.com/problems/word-ladder/

题目原文

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.

题目大意

给出开始单词、结束单词以及一个词典，求出从开始单词变换到结束单词所需要经过的词数，每次变换只能改变一个字母（若找不到这样的变化，返回0）

解题思路

根据开始单词，变换其中任意一个位置的字母，看新生成的单词是否在字典中出现，若新生成的单词等于结束单词，则结束，否则，继续变换新生成的单词

代码

class Solution(object):
    def ladderLength(self, beginWord, endWord, wordList):
        """
        :type beginWord: str
        :type endWord: str
        :type wordList: Set[str]
        :rtype: int
        """
        length = 2
        words = set([beginWord])
        wordList.discard(beginWord)
        wordList.discard(endWord)

        while words:
            newwords = set()
            for word in words:
                for i in range(len(word)):
                    for c in "abcdefghijklmnopqrstuvwxyz":
                        new_word = word[:i] + c + word[i + 1:]
                        if new_word == endWord:
                            return length
                        if new_word in wordList:
                            newwords.add(new_word)
            words = newwords
            length += 1
            wordList -= words
        return 0