leetcode 721. Accounts Merge 根据邮件合并账户 + 并查集Union Find

缺乏、安全感 2022-06-03 08:55 181阅读 0赞

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 \[1, 30\].

本体是通过并查集完成分类的，很棒的做法

建议和[leetcode 685. Redundant Connection II 并查集Union Find][leetcode 685. Redundant Connection II _Union Find] 一起学习

代码如下：

#include <iostream>
    #include <vector>
    #include <map>
    #include <set>
    #include <queue>
    #include <stack>
    #include <string>
    #include <climits>
    #include <algorithm>
    #include <sstream>
    #include <functional>
    #include <bitset>
    #include <numeric>
    #include <cmath>
    #include <regex>
    
    using namespace std;
    
    
    
    class Solution
    {
    public:
        vector<vector<string>> accountsMerge(vector<vector<string>>& accounts) 
        {
            vector<vector<string>> res;
            map<string, string> root;
            map<string, string> owner;
            map<string, set<string>> m;
            for (auto account : accounts)
            {
                for (int i = 1; i < account.size(); i++)
                {
                    root[account[i]] = account[i];
                    owner[account[i]] = account[0];
                }
            }
    
            for (auto account : accounts)
            {
                string p = find(account[1], root);
                for (int i = 2; i < account.size(); i++)
                {
                    root[find(account[i], root)] = p;
                }
            }
    
            for (auto account : accounts)
            {
                for (int i = 1; i < account.size(); i++)
                {
                    m[find(account[i], root)].insert(account[i]);
                }
            }
    
            for (auto a : m)
            {
                vector<string> one(a.second.begin(), a.second.end());
                one.insert(one.begin(), owner[a.first]);
                res.push_back(one);
            }
    
            return res;
        }
    
        string find(string s, map<string, string>& root)
        {
            if (root[s] == s)
                return s;
            else
                return find(root[s], root);
        }
    };

[leetcode 685. Redundant Connection II _Union Find]: http://blog.csdn.net/jackzhang_123/article/details/78891202