单词接龙
05/13/2025
题目描述
字典 wordList 中从单词 beginWord 到 endWord 的 转换序列 是一个按下述规格形成的序列 beginWord -> s1 -> s2 -> ... -> sk:
每一对相邻的单词只差一个字母。
对于 1 <= i <= k 时,每个 si 都在 wordList 中。注意, beginWord 不需要在 wordList 中。
sk == endWord
给你两个单词 beginWord 和 endWord 和一个字典 wordList ,返回 从 beginWord 到 endWord 的 最短转换序列 中的 单词数目 。如果不存在这样的转换序列,返回 0 。
示例 1:
输入:beginWord = "hit", endWord = "cog", wordList = ["hot","dot","dog","lot","log","cog"]
输出:5
解释:一个最短转换序列是 "hit" -> "hot" -> "dot" -> "dog" -> "cog", 返回它的长度 5。
示例 2:
输入:beginWord = "hit", endWord = "cog", wordList = ["hot","dot","dog","lot","log"]
输出:0
解释:endWord "cog" 不在字典中,所以无法进行转换。
提示:
1 <= beginWord.length <= 10
endWord.length == beginWord.length
1 <= wordList.length <= 5000
wordList[i].length == beginWord.length
beginWord、endWord 和 wordList[i] 由小写英文字母组成
beginWord != endWord
wordList 中的所有字符串 互不相同
本人题解
function ladderLength(beginWord: string, endWord: string, wordList: string[]): number {
if (!wordList.includes(endWord)) {
return 0;
}
const wordMap: Record<string, string[]> = {
[beginWord]: []
}
function setWordMap(word, nextWord, reverse = true) {
let curWordMap = wordMap[word];
if (!curWordMap) {
curWordMap = [];
wordMap[word] = curWordMap;
}
curWordMap.push(nextWord);
if (reverse) {
setWordMap(nextWord, word, false);
}
}
for (let i = 0; i < wordList.length; i++) {
const curWord = wordList[i];
if (isOneEditDistance(beginWord, curWord)) {
setWordMap(beginWord, curWord);
}
for (let j = i + 1; j < wordList.length; j++) {
const nextWord = wordList[j];
if (isOneEditDistance(curWord, nextWord)) {
setWordMap(curWord, nextWord);
}
}
}
return breadthFirstSearch(wordMap, beginWord, endWord);
};
function breadthFirstSearch(wordMap: Record<string, string[]>, beginWord: string, endWord: string): number {
let queues: string[][] = [[beginWord]];
let usedWords: string[] = [beginWord];
while(true) {
let _queues: string[][] = [];
for (let i = 0; i < queues.length; i++) {
const curQueue = queues[i];
const curWord = curQueue[curQueue.length - 1];
const nextWords = wordMap[curWord];
if (nextWords.includes(endWord)) {
return curQueue.length + 1;
}
_queues.push(...nextWords.filter(word => !usedWords.includes(word)).map(word => {
usedWords.push(word);
return [...curQueue, word]
}));
}
queues = _queues;
if (queues.length === 0) {
return 0;
}
}
}
function isOneEditDistance(word1: string, word2: string): boolean {
if (word1.length !== word2.length) {
return false;
}
let count = 0;
for (let i = 0; i < word1.length; i++) {
if (word1[i] !== word2[i]) {
count++;
}
}
return count === 1;
}
分析
先构建一个单词图,然后进行广度优先搜索。