Description
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
accountswill 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].
Solution
看了题目之后，原本美美地开始用HashMap写，后来发现是大坑，本题目corner case很多，考虑下面几种情况：
- [["n"]] // empty account
- [["n", "e0", "e1", "e01"]] // duplicate email in the same account
- [["n", "e0", "e1"], ["n0", "e1"]] // duplicate email in different account
- [["john", "e0", "e1"], ["john", "e2", "e3"], ["john", "e1", "e2"] // these three should be merged
对于最后一个case，我们可以看到其实这三个John是同一个人，但是贪婪算法遍历完前两个John，还是认为其是两个不同的人，当遍历第三个John时，就直接加到第一个John中了，而没有同时把第二个John加进来。
所以简单用HashMap不能解决此问题，必须采用图论的思想，使用DFS或者UnionFind。
DFS
首先build一个email之间的有向图，然后dfs即可。
class Solution {
public List<List<String>> accountsMerge(List<List<String>> accounts) {
Map<String, String> email2Name = new HashMap<>();
Map<String, Set<String>> graph = new HashMap<>(); // email to neighbors
// build graph
for (List<String> account : accounts) {
String name = account.get(0);
for (int i = 1; i < account.size(); ++i) {
String email = account.get(i);
email2Name.put(email, name);
if (!graph.containsKey(email)) {
graph.put(email, new HashSet<>());
}
if (i == 1) {
continue;
}
// add double directed edge between current email and the first email in account
String firstEmail = account.get(1);
graph.get(firstEmail).add(email);
graph.get(email).add(firstEmail);
}
}
// dfs
List<List<String>> res = new ArrayList<>();
Set<String> visitedEmails = new HashSet<>();
for (String email : graph.keySet()) {
if (!visitedEmails.add(email)) {
continue;
}
List<String> list = new ArrayList<>();
dfs(graph, email, list, visitedEmails);
// sort emails and insert name
Collections.sort(list);
list.add(0, email2Name.get(email));
res.add(list);
}
return res;
}
public void dfs(Map<String, Set<String>> graph, String node
, List<String> list, Set<String> visited) {
list.add(node);
for (String neighbor : graph.get(node)) {
if (visited.add(neighbor)) {
dfs(graph, neighbor, list, visited);
}
}
}
}
Union-find, time O(A * log A) or O(A), space O(A), A is total email count
build一个无向图，然后Union find。
用HashMap存储email2Eid，可以天然对email进行去重。
class Solution {
public List<List<String>> accountsMerge(List<List<String>> accounts) {
UnionFind uf = new UnionFind();
Map<String, Integer> email2Eid = new HashMap<>();
Map<String, String> email2Name = new HashMap<>();
int id = 0; // self-increase id
for (List<String> account : accounts) {
String name = account.get(0);
for (int i = 1; i < account.size(); ++i) {
String email = account.get(i);
if (!email2Eid.containsKey(email)) {
email2Eid.put(email, id++); // id should increase here!!!
email2Name.put(email, name);
}
// add an edge from current email to the first email of current account
uf.union(email2Eid.get(email), email2Eid.get(account.get(1)));
}
}
Map<Integer, List<String>> pidToAccount = new HashMap<>();
for (Map.Entry<String, Integer> entry : email2Eid.entrySet()) {
String email = entry.getKey();
int eid = entry.getValue();
int pid = uf.find(eid);
if (!pidToAccount.containsKey(pid)) {
List<String> account = new ArrayList<>();
account.add(email2Name.get(email)); // don't forget the name
pidToAccount.put(pid, account);
}
pidToAccount.get(pid).add(email);
}
for (List<String> account : pidToAccount.values()) {
Collections.sort(account); // sort finally
}
return new ArrayList<>(pidToAccount.values());
}
class UnionFind {
int[] parent;
public UnionFind() {
parent = new int[10000]; // at most 10000 emails
for (int i = 0; i < parent.length; ++i) {
parent[i] = i;
}
}
public int find(int x) {
if (parent[x] != x) {
parent[x] = find(parent[x]);
}
return parent[x];
}
public void union(int x, int y) {
parent[find(x)] = parent[find(y)];
}
}
}
