Question

721. 账户合并

English Version

题目描述

给定一个列表 accounts，每个元素 accounts[i] 是一个字符串列表，其中第一个元素 accounts[i][0] 是 _名称 (name)_，其余元素是 _emails _表示该账户的邮箱地址。

现在，我们想合并这些账户。如果两个账户都有一些共同的邮箱地址，则两个账户必定属于同一个人。请注意，即使两个账户具有相同的名称，它们也可能属于不同的人，因为人们可能具有相同的名称。一个人最初可以拥有任意数量的账户，但其所有账户都具有相同的名称。

合并账户后，按以下格式返回账户：每个账户的第一个元素是名称，其余元素是 按字符 ASCII 顺序排列 的邮箱地址。账户本身可以以 任意顺序 返回。

 

示例 1：

输入：accounts = [["John", "johnsmith@mail.com", "john00@mail.com"], ["John", "johnnybravo@mail.com"], ["John", "johnsmith@mail.com", "john_newyork@mail.com"], ["Mary", "mary@mail.com"]]
输出：[["John", 'john00@mail.com', 'john_newyork@mail.com', 'johnsmith@mail.com'],  ["John", "johnnybravo@mail.com"], ["Mary", "mary@mail.com"]]
解释：
第一个和第三个 John 是同一个人，因为他们有共同的邮箱地址 "johnsmith@mail.com"。 
第二个 John 和 Mary 是不同的人，因为他们的邮箱地址没有被其他帐户使用。
可以以任何顺序返回这些列表，例如答案 [['Mary'，'mary@mail.com']，['John'，'johnnybravo@mail.com']，
['John'，'john00@mail.com'，'john_newyork@mail.com'，'johnsmith@mail.com']] 也是正确的。

示例 2：

输入：accounts = [["Gabe","Gabe0@m.co","Gabe3@m.co","Gabe1@m.co"],["Kevin","Kevin3@m.co","Kevin5@m.co","Kevin0@m.co"],["Ethan","Ethan5@m.co","Ethan4@m.co","Ethan0@m.co"],["Hanzo","Hanzo3@m.co","Hanzo1@m.co","Hanzo0@m.co"],["Fern","Fern5@m.co","Fern1@m.co","Fern0@m.co"]]
输出：[["Ethan","Ethan0@m.co","Ethan4@m.co","Ethan5@m.co"],["Gabe","Gabe0@m.co","Gabe1@m.co","Gabe3@m.co"],["Hanzo","Hanzo0@m.co","Hanzo1@m.co","Hanzo3@m.co"],["Kevin","Kevin0@m.co","Kevin3@m.co","Kevin5@m.co"],["Fern","Fern0@m.co","Fern1@m.co","Fern5@m.co"]]

 

提示：

• 1 <= accounts.length <= 1000
• 2 <= accounts[i].length <= 10
• 1 <= accounts[i][j].length <= 30
• accounts[i][0] 由英文字母组成
• accounts[i][j] (for j > 0) 是有效的邮箱地址

解法

方法一

class Solution:
  def accountsMerge(self, accounts: List[List[str]]) -> List[List[str]]:
    def find(x):
      if p[x] != x:
        p[x] = find(p[x])
      return p[x]

    n = len(accounts)
    p = list(range(n))
    email_id = {}
    for i, account in enumerate(accounts):
      name = account[0]
      for email in account[1:]:
        if email in email_id:
          p[find(i)] = find(email_id[email])
        else:
          email_id[email] = i
    mp = defaultdict(set)
    for i, account in enumerate(accounts):
      for email in account[1:]:
        mp[find(i)].add(email)

    ans = []
    for i, emails in mp.items():
      t = [accounts[i][0]]
      t.extend(sorted(emails))
      ans.append(t)
    return ans

class Solution {
  private int[] p;

  public List<List<String>> accountsMerge(List<List<String>> accounts) {
    int n = accounts.size();
    p = new int[n];
    for (int i = 0; i < n; ++i) {
      p[i] = i;
    }
    Map<String, Integer> emailId = new HashMap<>();
    for (int i = 0; i < n; ++i) {
      List<String> account = accounts.get(i);
      String name = account.get(0);
      for (int j = 1; j < account.size(); ++j) {
        String email = account.get(j);
        if (emailId.containsKey(email)) {
          p[find(i)] = find(emailId.get(email));
        } else {
          emailId.put(email, i);
        }
      }
    }
    Map<Integer, Set<String>> mp = new HashMap<>();
    for (int i = 0; i < n; ++i) {
      List<String> account = accounts.get(i);
      for (int j = 1; j < account.size(); ++j) {
        String email = account.get(j);
        mp.computeIfAbsent(find(i), k -> new HashSet<>()).add(email);
      }
    }
    List<List<String>> res = new ArrayList<>();
    for (Map.Entry<Integer, Set<String>> entry : mp.entrySet()) {
      List<String> t = new LinkedList<>();
      t.addAll(entry.getValue());
      Collections.sort(t);
      t.add(0, accounts.get(entry.getKey()).get(0));
      res.add(t);
    }
    return res;
  }

  private int find(int x) {
    if (p[x] != x) {
      p[x] = find(p[x]);
    }
    return p[x];
  }
}

class Solution {
public:
  vector<int> p;

  vector<vector<string>> accountsMerge(vector<vector<string>>& accounts) {
    int n = accounts.size();
    p.resize(n);
    for (int i = 0; i < n; ++i) p[i] = i;
    unordered_map<string, int> emailId;
    for (int i = 0; i < n; ++i) {
      auto account = accounts[i];
      auto name = account[0];
      for (int j = 1; j < account.size(); ++j) {
        string email = account[j];
        if (emailId.count(email))
          p[find(i)] = find(emailId[email]);
        else
          emailId[email] = i;
      }
    }
    unordered_map<int, unordered_set<string>> mp;
    for (int i = 0; i < n; ++i) {
      auto account = accounts[i];
      for (int j = 1; j < account.size(); ++j) {
        string email = account[j];
        mp[find(i)].insert(email);
      }
    }
    vector<vector<string>> ans;
    for (auto& [i, emails] : mp) {
      vector<string> t;
      t.push_back(accounts[i][0]);
      for (string email : emails) t.push_back(email);
      sort(t.begin() + 1, t.end());
      ans.push_back(t);
    }
    return ans;
  }

  int find(int x) {
    if (p[x] != x) {
      p[x] = find(p[x]);
    }
    return p[x];
  }
};