one set [visited] record visited words in last layer, one set [current] record words in the current layer, one set [dict] record words in dict(candidates for the ladder) change each letter of the words in [visited] to see whether dict contains the new word, if yes, remove the word ...
classSolution {public:intladderLength(stringbeginWord,stringendWord, vector<string>&wordList) { unordered_set<string>dict(wordList.begin(), wordList.end()); queue<string>q; q.push(beginWord);intladder =1;while(!q.empty()) {intn =q.size();for(inti =0; i < n; i++) {stringcur =q...
第一次做; 融合了力扣题解的通用状态哈希表以及LeetCode题解的双向bfs; 核心: 双向bfs并不是同时双向, 实际上是这次从beginWord方向bfs,下次从endWord方向bfs; 代码中实现双向遍历的方式是while循环中的第一个if语句; 代码里没法实现真正的同时双向bfs, 我想了想, 同不同时并不重要, 重要的是从两端进行bfs; ...
优化手段map<string,int> words; 表示单词(string)是通过改第几位得到的。class Solution { public: int ladderLength(string beginWord, string endWord, vector<string>& wordList) { set<string> dicts(wordList.begin(),wordList.end()); // I change the unordered_set to set, and it takes more tim...
LeetCode: Word Ladder II [127] 【题目】 Given two words (startandend), and a dictionary, find all shortest transformation sequence(s) fromstarttoend, such that: Only one letter can be changed at a time Each intermediate word must exist in the dictionary...
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 { ...
importcopyclass Solution:defladderLength(self,beginWord:str,endWord:str,wordList:List[str])->int:ifendWord not in wordList:return0alphabet=[chr(ord('a')+i)fori inrange(26)]defdistance_one(string):string=list(string)ret=[]fori inrange(len(string)):cache=copy.copy(string)forj in alphabe...
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. ...
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
[LeetCode] 126. Word Ladder II Problem Given two words (start and end), and a dictionary, find all shortest transformation sequence(s) from start to end, such that: Only one letter can be changed at a time Each intermediate word must exist in the dictionary...