721. 账户合并(并查集/深搜/广搜)

题目大意: 同一个昵称可能不是同一个人,也可能是同一个人,但不同的昵称肯定不是同一个人.按照题意将同一个人的邮箱地址链接起来

并查集:

　　对对应的邮箱地址进行并查集操作,如果存在有交集的邮箱地址,则两个列表肯定归属于同一个人,将他们连接起来.

class Solution {
public:
    int find(int idr, vector<int>& u) {
        while (idr != u[idr]) idr = u[idr];
        return idr;
    }
    void join(int idr1, int idr2, vector<int>& u) {
        int f1 = find(idr1, u);
        int f2 = find(idr2, u);
        u[f2] = f1;
    }
    map<string, int> hashmap1;
    map<int, vector<vector<string>>> hashmap2;
    vector<vector<string>> res; //返回向量
    vector<vector<string>> accountsMerge(vector<vector<string>>& accounts) {
        vector<int> u(accounts.size()); 
        for (int i = 0; i < u.size(); i++) u[i] = i; //初始化并查集
        for (int k = 0; k < accounts.size(); k++) { //查找,逻辑连接
            for (int i = 1;i<accounts[k].size();i++) {
                auto addr = accounts[k][i];
                auto iter = hashmap1.find(addr);
                if (iter == hashmap1.end()) hashmap1[addr] = k;
                else join(hashmap1[addr], k, u);
            }
        }

        //打印并查集情况
        //for (int i = 0; i < u.size(); i++) {
        //    cout << u[i] << " ";
        //    for (auto val : accounts[i]) cout << val << " ";
        //    cout << endl;
        //}

        for (int i = 0; i < accounts.size(); i++) { //物理连接
            int f = find(u[i], u); //对顶层节点进行分类
            auto iter = hashmap2.find(f);
            if (iter == hashmap2.end()) { //如果还不存在先创建一个
                hashmap2[f] = {};
            }
            hashmap2[f].push_back(accounts[i]);
        }
        for (auto iter = hashmap2.begin(); iter != hashmap2.end(); iter++) {
            set<string> map;
            for (auto list : iter->second) {
                for (int i = 1;i<list.size();i++) {
                    string val = list[i];
                    auto itr = map.find(val);
                    if (itr == map.end()) map.insert(val);
                }
            }
            if (map.empty()) continue;
            vector<string> temp(map.begin(), map.end());
            sort(temp.begin(), temp.end());
            temp.insert(temp.begin(), accounts[iter->first][0]);
            res.push_back(temp);
        }
        return res;
    }
};