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123 lines
4.9 KiB
Markdown
123 lines
4.9 KiB
Markdown
# [127. Word Ladder](https://leetcode.com/problems/word-ladder/)
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# 思路
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词句阶梯: 给了我们一个单词字典，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。
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要求最短路径长度，首先应该想到BFS。类似迷宫里面的最短路求法，迷宫每个点有上下左右四个方向可以走，而这里有26个字母，就是二十六个方向可以走，本质上没有区别。
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另外由于需要频繁判断某个单词是否在字典中，所以我们用hashSet记录词典（代码中的`wordSet`）以快速进行判断。
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## 思路一
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采用递归的BFS。首先我们用一个词语数组`beginWords`来记录开始的词语，初始化为一个元素`beginWord`，此外类似走迷宫的题目我们还需要一个`visited`来记录当前词语是否被访问过，这里我们用hashSet来定义`visited`。递归出口是`beginWords`为空。
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## 思路二、思路一改进
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仔细分析我们可以发现了个hashSet`wordSet`和`visited`其实在功能上是有冗余的: 如果我们每访问一次`wordSet`里的词然后就把它从`wordSet`中去掉，这样不就不需要`visited`了吗？亲测会比思路一快很多。
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## 思路三
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思路二的非递归版本。此时就不需要`beginWords`了而需要一个队列。
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# C++
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## 思路一
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``` C++
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class Solution {
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private:
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int BFS(vector<string>beginWords, unordered_set<string>&wordSet,
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unordered_set<string>&visited, string &endWord){
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vector<string>next_beginWords;
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string word;
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for(int i = 0; i < beginWords.size(); i++){
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for(int j = 0; j < endWord.size(); j++){
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word = beginWords[i];
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if(word == endWord) return 1;
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for(char k = 'a'; k <= 'z'; k++){
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word[j] = k;
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if(!wordSet.count(word) || visited.count(word)) continue;
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visited.insert(word);
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next_beginWords.push_back(word);
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}
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}
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}
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if(next_beginWords.empty()) return 0; // 递归出口
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return BFS(next_beginWords, wordSet, visited, endWord) + 1;
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}
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public:
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int ladderLength(string beginWord, string endWord, vector<string>& wordList) {
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unordered_set<string>wordSet(wordList.begin(), wordList.end());
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unordered_set<string>visited{beginWord};
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if(wordSet.count(endWord) != 1) return 0;
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vector<string>beginWords{beginWord};
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return BFS(beginWords, wordSet, visited, endWord);
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}
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};
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```
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## 思路二
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``` C++
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class Solution {
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private:
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int BFS(vector<string>beginWords, unordered_set<string>&wordSet, string &endWord){
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vector<string>next_beginWords;
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string word;
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for(int i = 0; i < beginWords.size(); i++){
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for(int j = 0; j < endWord.size(); j++){
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word = beginWords[i];
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if(word == endWord) return 1;
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char tmp_c = word[j];
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for(char k = 'a'; k <= 'z'; k++){
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word[j] = k;
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// tmp_c == k即新旧单词相同
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if(!wordSet.count(word) || tmp_c == k) continue;
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wordSet.erase(word);
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next_beginWords.push_back(word);
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}
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}
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}
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if(next_beginWords.empty()) return 0; // 递归出口
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return BFS(next_beginWords, wordSet, endWord) + 1;
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}
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public:
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int ladderLength(string beginWord, string endWord, vector<string>& wordList) {
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unordered_set<string>wordSet(wordList.begin(), wordList.end());
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if(wordSet.count(endWord) != 1) return 0;
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vector<string>beginWords{beginWord};
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return BFS(beginWords, wordSet, endWord);
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}
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};
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```
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## 思路三
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``` C++
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class Solution {
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public:
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int ladderLength(string beginWord, string endWord, vector<string>& wordList) {
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unordered_set<string> wordSet(wordList.begin(), wordList.end());
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if (!wordSet.count(endWord)) return 0;
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queue<string> q;
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q.push(beginWord);
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int res = 0;
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while (!q.empty()) {
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for (int k = q.size(); k > 0; --k) {
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string word = q.front(); q.pop();
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if (word == endWord) return res + 1;
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for (int i = 0; i < word.size(); ++i) {
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string newWord = word;
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for (char k = 'a'; k <= 'z'; ++k) {
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newWord[i] = k;
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if (wordSet.count(newWord) && newWord != word) {
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q.push(newWord);
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wordSet.erase(newWord);
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}
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}
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}
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}
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++res;
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}
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return 0;
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}
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};
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```
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