Can you solve this real interview question? Word Ladder - A transformation sequence from word beginWord to word endWord using a dictionary wordList is a sequence of words beginWord -> s1 -> s2 -> ... -> sk such that: * Every adjacent pair of words diff
However, I implemented this BFS but got a TLE. So I adapted adouble-direction BFSwhich may half the running time, and is accepted by the leetcode judge. The algorithm can go as follows. Let start_front be the list of edge words of BFS paths from the start word Let end_front be the...
1classSolution {2public:3vector<vector<string>> findLadders(stringstart,stringend, unordered_set<string> &dict) {4//用于存状态的一些变量5queue<string>start_q, end_q;6unordered_map<string,int>depth_from_start, depth_from_end;7unordered_set<string>meet;8depth_from_start[start] =0;9depth_...
} public int ladderLength(String beginWord, String endWord, Set<String> wordList) { Queue<Word> queue = new LinkedList<Word>(); queue.offer(new Word(beginWord, 1)); wordList.add(endWord); while (!queue.isEmpty()) { Word w = queue.poll(); if (w.word.equals(endWord)) { return w...
func ladderLength(beginWord string, endWord string, wordList []string) int { // 先把开始单词放入单词列表中,方便后续使用下标处理 wordList = append(wordList, beginWord) startIndex := len(wordList) - 1 // 找到结束单词在单词列表中的下标 endIndex := -1 for i, word := range wordList { ...
res.remove(i); //删除之后因为i会加一,所以i需要往前退一格 i--; } } return res; } private void dfs(String beginWord, String endWord, List<String> wordList, List<List<String>> res, List<String> item, boolean[] visited){ if(beginWord.equals(endWord)){ ...
LeetCode Top Interview Questions 127. Word Ladder (Java版; Medium) 题目描述 AI检测代码解析 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: ...
LeetCode 0127. Word Ladder单词接龙【Medium】【Python】【BFS】 Problem LeetCode Given two words (beginWordandendWord), and a dictionary's word list, find the length of shortest transformation sequence frombeginWordtoendWord, such that: Only one letter can be changed at a time. ...
Word Ladder 一个单词字典,里面有一系列很相似的单词,然后给了一个起始单词和一个结束单词,每次变换只能改变一个单词,并且中间过程的单词都必须是单词字典中的单词,让我们求出最短的变化序列的长度。 https://leetcode.com/problems/word-ladder/ Input:beginWord="hit",endWord="cog",wordList=["hot","dot",...
Word Ladder II 一开始以为做了I之后,用一样的思路解决II应该会容易很多。但做了才发现,II对效率的要求更高,而且这使得这一题成为了LeetCode上通过率最低的题之一。 先讲一下自己做的时候的思路,和I相似,用图的思想,只不过是该用DFS(深度优先搜索),从起点出发,对字典中的每一个可达字符串进行DFS并记录路径...