721 Accounts Merge

1. Question

Given a listaccounts, each elementaccounts[i]is a list of strings, where the first elementaccounts[i][0]is aname, and the rest of the elements areemailsrepresenting 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:
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Input:
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accounts = [["John", "[email protected]", "[email protected]"], ["John", "[email protected]"], ["John", "[email protected]", "[email protected]"], ["Mary", "[email protected]"]]
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Output:
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Explanation:
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The first and third John's are the same person as they have the common email "[email protected]".
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The second John and Mary are different people as none of their email addresses are used by other accounts.
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We could return these lists in any order, for example the answer [['Mary', '[email protected]'], ['John', '[email protected]'],
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['John', '[email protected]', '[email protected]', '[email protected]']] would still be accepted.
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Note:
The length ofaccountswill be in the range[1, 1000].
The length ofaccounts[i]will be in the range[1, 10].
The length ofaccounts[i][j]will be in the range[1, 30].

2. Implementation

(1) DFS
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class Solution {
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public List<List<String>> accountsMerge(List<List<String>> accounts) {
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List<List<String>> res = new ArrayList<>();
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if (accounts == null || accounts.size() == 0) {
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return res;
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}
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Map<String, String> emailToName = new HashMap<>();
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Map<String, Set<String>> graph = new HashMap<>();
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Set<String> emails = new HashSet<>();
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for (List<String> account : accounts) {
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String name = account.get(0);
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for (int i = 1; i < account.size(); i++) {
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String email = account.get(i);
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emails.add(email);
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emailToName.put(email, name);
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graph.putIfAbsent(email, new HashSet<>());
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// Build edge
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if (i != 1) {
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graph.get(account.get(i - 1)).add(email);
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graph.get(email).add(account.get(i - 1));
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}
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}
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}
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Set<String> visited = new HashSet<>();
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for (String email : emails) {
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if (!visited.contains(email)) {
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visited.add(email);
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List<String> buffer = new ArrayList<>();
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buffer.add(email);
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findConnectedComponentByDFS(email, graph, visited, buffer);
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Collections.sort(buffer);
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buffer.add(0, emailToName.get(email));
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res.add(buffer);
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}
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}
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return res;
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}
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public void findConnectedComponentByDFS(String email, Map<String, Set<String>> graph, Set<String> visited, List<String> buffer) {
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for (String nextEmail : graph.get(email)) {
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if (!visited.contains(nextEmail)) {
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visited.add(nextEmail);
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buffer.add(nextEmail);
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findConnectedComponentByDFS(nextEmail, graph, visited, buffer);
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}
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}
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}
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}
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3. Time & Space Complexity

DFS: 时间复杂度O(k * nlogn) ,n是输入的每个email list平均长度, k是email list的个数, 空间复杂度O(nk)