原题链接在这里：https://leetcode.com/problems/accounts-merge/
题目：
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", "[email protected]", "[email protected]"], ["John", "[email protected]"], ["John", "[email protected]", "[email protected]"], ["Mary", "[email protected]"]]
Output: [["John", '[email protected]', '[email protected]', '[email protected]'], ["John", "[email protected]"], ["Mary", "[email protected]"]]
Explanation:
The first and third John's are the same person as they have the common email "[email protected]".
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', '[email protected]'], ['John', '[email protected]'],
['John', '[email protected]', '[email protected]', '[email protected]']] would still be accepted.
Note:
- The length of
accountswill 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].
题解：
Each account has an email cluster. Need to find one to represent the cluster. Here use the first email address as represent.
Find out the parent of represent, and assign it as the parent for all other email parents' parent.
Check each represent's parent, add all emails into this union.
Time Complexity: O(nlogn). n is number of email address.
Space: O(n).
AC Java:
class Solution {
HashMap<String, String> parent;
public List<List<String>> accountsMerge(List<List<String>> accounts) {
parent = new HashMap<>();
HashMap<String, String> ownersMap = new HashMap<>();
HashMap<String, Set<String>> unions = new HashMap<>();
List<List<String>> res = new ArrayList<List<String>>();
for(List<String> item : accounts){
for(int i = 1; i<item.size(); i++){
parent.put(item.get(i), item.get(i));
ownersMap.put(item.get(i), item.get(0));
}
}
for(List<String> item : accounts){
String p = find(item.get(1));
for(int i = 2; i<item.size(); i++){
parent.put(find(item.get(i)), p);
}
}
for(List<String> item : accounts){
String p = find(item.get(1));
unions.putIfAbsent(p, new HashSet<String>());
for(int i = 1; i<item.size(); i++){
unions.get(p).add(item.get(i));
}
}
for(String represent : unions.keySet()){
String owner = ownersMap.get(represent);
List<String> union = new ArrayList(unions.get(represent));
Collections.sort(union);
union.add(0, owner);
res.add(union);
}
return res;
}
private String find(String i){
if(i != parent.get(i)){
parent.put(i, find(parent.get(i)));
}
return parent.get(i);
}
}