【leetcode】(python)127. Word Ladder单词阶梯

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最新推荐文章于 2020-10-26 16:17:55 发布

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探讨如何寻找两个单词间最短的转换序列，通过仅改变一个字母的方式，确保每一步转换后的新单词都存在于给定的字典中。利用队列实现广度优先搜索算法，高效解决问题。

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127. Word Ladder Medium

Description

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

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.

题意

给定两个单词（beginWord和endWord）和一个字典的单词列表，找到从beginWord到endWord的最短变换序列的长度。每次只能改动上一单词的一个字母，并且改动后的单词必须存在于单词列表中，beginWord不是转换后的单词，直接计入长度。

解题思路

这题借助队列，将转换后的单词存入队列中，然后再遍历单词列表，如果满足题意，最后结果就是队列的长度。

code

class Solution:
    def ladderLength(self, beginWord: str, endWord: str, wordList: List[str]) -> int:
        if endWord not in wordList:
            return 0
        wordset = set(wordList)
        Q = collections.deque()
        Q.append((beginWord,1))
        while Q:
            word, length = Q.popleft()
            if word == endWord:
                return length
            for i in range(len(word)):
                for j in 'abcdefghijklmnopqrstuvwxyz':
                    newWord = word[:i] + j + word[i+1:]
                    if newWord != word and newWord in wordset:
                        wordset.remove(newWord)
                        Q.append((newWord,length+1))
        return 0