[leetcode] 127. Word Ladder

最新推荐文章于 2024-09-10 03:45:52 发布

原创最新推荐文章于 2024-09-10 03:45:52 发布 · 308 阅读

CC 4.0 BY-SA版权

本文介绍了一种算法，用于找到两个单词之间的最短转换路径，仅允许每次更改一个字母，且所有转换后的单词必须在给定的词典中存在。通过使用双向队列进行广度优先搜索（BFS），该算法能在O(n)时间内找到从开始单词到目标单词的最短路径长度。

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 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 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.

解法

BFS, 使用双向队列储存每次转换的结果
参考: https://blog.youkuaiyun.com/fuxuemingzhu/article/details/82903681
Time: O(n)
Space: O(n)

代码

class Solution:
    def ladderLength(self, beginWord, endWord, wordList):
        """
        :type beginWord: str
        :type endWord: str
        :type wordList: List[str]
        :rtype: int
        """
        # edge case
        if endWord not in wordList:
            return 0
        # bfs
        wordSet = set(wordList)
        queue = collections.deque()
        queue.append((beginWord, 1))
        while queue:
            word, length = queue.popleft()
            if word == endWord:
                return length
            for i in range(len(word)):
                for ch in 'abcdefghijklmnopqrstuvwxyz':
                    newWord = word[:i] + ch + word[i+1:]
                    if newWord in wordSet and newWord != word:
                        queue.append((newWord, length+1))
                        wordSet.remove(newWord)
        return 0