LeetCode|Accounts Merge

Accounts Merge

Given a list accounts, each element accounts[i] is a list of strings, where the first element accounts[i][0] is a name, and the rest of the elements are emails representing emails of the account.

Now, we would like to merge these accounts. Two accounts definitely belong to the same person if there is some email that is common to both accounts. Note that even if two accounts have the same name, they may belong to different people as people could have the same name. A person can have any number of accounts initially, but all of their accounts definitely have the same name.

After merging the accounts, return the accounts in the following format: the first element of each account is the name, and the rest of the elements are emails in sorted order. The accounts themselves can be returned in any order.

Example 1:

Input: 
accounts = [["John", "johnsmith@mail.com", "john00@mail.com"], ["John", "johnnybravo@mail.com"], ["John", "johnsmith@mail.com", "john_newyork@mail.com"], ["Mary", "mary@mail.com"]]
Output: [["John", 'john00@mail.com', 'john_newyork@mail.com', 'johnsmith@mail.com'],  ["John", "johnnybravo@mail.com"], ["Mary", "mary@mail.com"]]
Explanation: 
The first and third John's are the same person as they have the common email "johnsmith@mail.com".
The second John and Mary are different people as none of their email addresses are used by other accounts.
We could return these lists in any order, for example the answer [['Mary', 'mary@mail.com'], ['John', 'johnnybravo@mail.com'], 
['John', 'john00@mail.com', 'john_newyork@mail.com', 'johnsmith@mail.com']] would still be accepted.

Note:

• The length of accounts will be in the range [1, 1000].
• The length of accounts[i] will be in the range [1, 10].
• The length of accounts[i][j] will be in the range [1, 30].

class Solution {
public:
    vector<vector<string> > accountsMerge(vector<vector<string>>& acc) {
        unordered_map<string, int> m;  // eamil 2 id
        int id = acc.size();  // 0~acc.size()-1 for name, otherwise for email
        vector<vector<string> > res;
        parent = vector<int>(10000);
        rank =  vector<int>(10000, 0);
        /* 一个账户可能连接了好几个accounts, 所以应该先做完并查集的操作 */
        for(int i = 0; i < acc.size(); ++i){
            parent[i] = i;
            for(int j = 1; j < acc[i].size(); ++j){
                auto it = m.find(acc[i][j]);
                if(it == m.end()){
                    m[acc[i][j]] = id;
                    parent[id++] = i;
                } else{
                    int id_val = it->second;
                    union_set(i, id_val);  // 如果一个acc下有重复email。那么要么跟之前的i'合并，要么跟i合并. (i'< i)
                }
            }
        }
        id = 0;  // id for res
        vector<int> name2id(10000, -1);
        for(int i = 0; i < acc.size(); ++i){
            int parent_id = find(i);
            if(name2id[parent_id] == -1){
                res.push_back(vector<string>(1, acc[i][0]));
                name2id[parent_id] = id++;
            }
            int pos = name2id[parent_id];
            for(int j = 1; j < acc[i].size(); ++j){
                auto it = m.find(acc[i][j]);
                if(it != m.end()){
                    res[pos].push_back(acc[i][j]);
                    m.erase(it);
                }
                
            }
        }
        for(int i = 0; i < res.size(); ++i){
            sort(res[i].begin()+1, res[i].end());
        } return res;
    }
private:
    vector<int> parent, rank;
    int find(int x){
        if(x != parent[x]) parent[x] = find(parent[x]);
        return parent[x];
    }
    void union_set(int x, int y){
        if((x=find(x)) == (y=find(y))) return;
        if(rank[x] > rank[y]) parent[y] = x;
        else{
            parent[x] = y;
            if(rank[x] == rank[y]) ++rank[y];
        }
    }
};