LWC 57:721. Accounts Merge

LWC 57:721. Accounts Merge

传送门:721. Accounts Merge

Problem:

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].

思路: 很暴力,简单的从头到尾慢慢添加新来的信息。首先,如果该用户在原先的数据库中不存在,则可以直接加入数据库。如果该用户存在与数据库,注意两种情况:

  • 同名, 但没有邮箱重复,说明不是同一个人,直接加入数据库。
  • 同名,且邮箱有重复,说明是同一个人,则把所有关联的邮箱进行合并即可。

代码如下:

    public List<List<String>> accountsMerge(List<List<String>> accounts) {
        Map<String, List<Set<String>>> mem = new HashMap<>();
        int n = accounts.size();
        for (int i = 0; i < n; ++i) {
            List<String> account = accounts.get(i);
            String name = account.get(0);
            if (mem.containsKey(name)) {
                List<Set<String>> mm = mem.get(name);
                int idx = -1;
                Set<String> hh = new HashSet<>();
                for (int k = 1; k < account.size(); ++k){
                    hh.add(account.get(k));
                }
                Set<Integer> mer = new HashSet<>();
                for (int k = 1; k < account.size(); ++k) {
                    String email = account.get(k);
                    for (int j = 0; j < mm.size(); ++j) {
                        if (mm.get(j).contains(email)) {
                            idx = j;
                            mer.add(idx);
                        }
                    }
                }

                if (mer.size() == 0) { // 同名,属于两个人
                    mem.computeIfAbsent(name, k -> new ArrayList<>()).add(hh);
                }
                else { // 属于同一个人
                    // 邮箱合并
                    List<Integer> merge = new ArrayList<>();
                    merge.addAll(mer);

                    mm.get(merge.get(0)).addAll(hh);
                    List<Set<String>> removes = new ArrayList<>();
                    for (int j = 1; j < merge.size(); ++j) {
                        mm.get(merge.get(0)).addAll(mm.get(merge.get(j)));
                        removes.add(mm.get(merge.get(j)));
                    }

                    // 删除之前的邮箱
                    for (Set<String> re : removes) {
                        mm.remove(re);
                    }
                }
            }
            else {
                // 数据库中不存在该用户,直接加入数据库
                Set<String> mails = new HashSet<>();
                for (int j = 1; j < account.size(); ++j) {
                    mails.add(account.get(j));
                }
                mem.computeIfAbsent(name, k -> new ArrayList<>()).add(mails);
            }
        }

        List<List<String>> ans = new ArrayList<>();
        for (String key : mem.keySet()) {
            List<Set<String>> val = mem.get(key);
            for (int i = 0; i < val.size(); ++i) {
                List<String> vv = new ArrayList<>();
                vv.addAll(val.get(i));
                Collections.sort(vv);
                List<String> tmp = new ArrayList<>();
                tmp.add(key);
                tmp.addAll(vv);
                ans.add(tmp);
            }
        }
        return ans;
    }   

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