[LeetCode] 721. Accounts Merge

lk20150415 发布于2019-08-16 13:48 / 364人阅读

Problem

Given a list accounts, each element accounts[i] is a list of strings, where the first element accountsi 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 accountsi will be in the range [1, 30].

Solution

class Solution {
    public List> accountsMerge(List> accounts) {
        //construct email graph
        Map parents = new HashMap<>();
        //save email-user relationships
        Map users = new HashMap<>();
        
        //initialize email graph, and save email-user map, btw
        for (List account: accounts) {
            for (int i = 1; i < account.size(); i++) {
                parents.put(account.get(i), account.get(i));
                users.put(account.get(i), account.get(0));
            }
        }
        
        //find parent and union to the first node"s parent in each group
        for (List account: accounts) {
            String parent = find(parents, account.get(1));
            for (int i = 2; i < account.size(); i++) {
                String curParent = find(parents, account.get(i));
                parents.put(curParent, parent);
            }
        }
        
        //loop through groups and save the unions
        Map> unions = new HashMap<>();
        for (List account: accounts) {
            String parent = find(parents, account.get(1));
            if (!unions.containsKey(parent)) {
                unions.put(parent, new TreeSet<>());
            }
            for (int i = 1; i < account.size(); i++) {
                unions.get(parent).add(account.get(i));
            }
        }
        
        //loop through unions map and save to result
        List> res = new ArrayList<>();
        for (String parent: unions.keySet()) {
            List emails = new ArrayList<>(unions.get(parent));
            emails.add(0, users.get(parent)); //put user name at start pos
            res.add(emails);
        }
        
        return res;
    }
    
    private String find(Map parents, String str) {
        if (!parents.containsKey(str)) return null;
        String parent = parents.get(str);
        if (parent.equals(str)) return parent;
        else return find(parents, parent);
    }
}

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