LeetCode 1258. 近义词句子（哈希+并查集+排序+回溯）

文章目录

- 1. 题目
- 2. 解题

1. 题目

给你一个近义词表 synonyms 和一个句子 text ， synonyms 表中是一些近义词对，你可以将句子 text 中每个单词用它的近义词来替换。

请你找出所有用近义词替换后的句子，按 字典序排序 后返回。

示例 1：
输入：
synonyms = [["happy","joy"],["sad","sorrow"],["joy","cheerful"]],
text = "I am happy today but was sad yesterday"
输出：
["I am cheerful today but was sad yesterday",
"I am cheerful today but was sorrow yesterday",
"I am happy today but was sad yesterday",
"I am happy today but was sorrow yesterday",
"I am joy today but was sad yesterday",
"I am joy today but was sorrow yesterday"]提示：
0 <= synonyms.length <= 10
synonyms[i].length == 2
synonyms[0] != synonyms[1]
所有单词仅包含英文字母，且长度最多为 10 。
text 最多包含 10 个单词，且单词间用单个空格分隔开。

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/synonymous-sentences
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

2. 解题

class dsu
{vector<int> f;
public:dsu(int n){f.resize(n);for(int i = 0; i < n; ++i)f[i] = i;}void merge(int a, int b){int fa = find(a);int fb = find(b);f[fa] = fb;}int find(int a){int origin = a;while(a != f[a])a = f[a];return f[origin] = a;}
};
class Solution {unordered_map<string, int> w_id;//单词 id映射unordered_map<int, string> id_w;//id 单词unordered_map<int, vector<string>> f_words;//近义词代表id, 近义词集合vector<string> ans;//答案
public:vector<string> generateSentences(vector<vector<string>>& synonyms, string text) {int i = 0;for(auto& s : synonyms){if(!w_id.count(s[0])){w_id[s[0]] = i;id_w[i++] = s[0];}if(!w_id.count(s[1])){w_id[s[1]] = i;id_w[i++] = s[1];}}int n = w_id.size(), i1, i2, f;//并查集找集合dsu u(n);for(auto& s : synonyms){i1 = w_id[s[0]];i2 = w_id[s[1]];u.merge(i1, i2);//近义词合并}for(i = 0; i  < n; ++i){f = u.find(i);//近义词代表的idf_words[f].push_back(id_w[i]);//加入集合}for(auto& fw : f_words)sort(fw.second.begin(), fw.second.end());//近义词排序vector<string> sentenceWords;//获取句子里的单词string w;for(int i = 0; i < text.size(); ++i){if(text[i] == ' ' || i == text.size()-1){if(i == text.size()-1) w += text[i];sentenceWords.push_back(w);w = "";}elsew += text[i];}string path;bt(sentenceWords, 0, path, u);//回溯生成句子return ans;}void bt(vector<string>& sentenceWords, int i, string& path, dsu& u){if(i == sentenceWords.size()){path.pop_back();//空格ans.push_back(path);return;}int size = path.size();if(!w_id.count(sentenceWords[i])){   //没有近义词path += sentenceWords[i]+" ";bt(sentenceWords, i+1, path, u);path.resize(size);//回溯}else{int f = u.find(w_id[sentenceWords[i]]);//有近义词，近义词的代表ffor(int j = 0; j < f_words[f].size(); ++j)//遍历近义词集合{path += f_words[f][j]+" ";bt(sentenceWords, i+1, path, u);path.resize(size);//回溯}}}
};