标签：Everyone logs int timestamp When Moment become friends log
原题链接在这里：https://leetcode.com/problems/the-earliest-moment-when-everyone-become-friends/
题目：
In a social group, there are N people, with unique integer ids from 0 to N-1.
We have a list of logs, where each logs[i] = [timestamp, id_A, id_B] contains a non-negative integer timestamp, and the ids of two different people.
Each log represents the time in which two different people became friends. Friendship is symmetric: if A is friends with B, then B is friends with A.
Let's say that person A is acquainted with person B if A is friends with B, or A is a friend of someone acquainted with B.
Return the earliest time for which every person became acquainted with every other person. Return -1 if there is no such earliest time.
Example 1:
Input: logs = [[20190101,0,1],[20190104,3,4],[20190107,2,3],[20190211,1,5],[20190224,2,4],[20190301,0,3],[20190312,1,2],[20190322,4,5]], N = 6 Output: 20190301 Explanation: The first event occurs at timestamp = 20190101 and after 0 and 1 become friends we have the following friendship groups [0,1], [2], [3], [4], [5]. The second event occurs at timestamp = 20190104 and after 3 and 4 become friends we have the following friendship groups [0,1], [2], [3,4], [5]. The third event occurs at timestamp = 20190107 and after 2 and 3 become friends we have the following friendship groups [0,1], [2,3,4], [5]. The fourth event occurs at timestamp = 20190211 and after 1 and 5 become friends we have the following friendship groups [0,1,5], [2,3,4]. The fifth event occurs at timestamp = 20190224 and as 2 and 4 are already friend anything happens. The sixth event occurs at timestamp = 20190301 and after 0 and 3 become friends we have that all become friends.
Note:
1 <= N <= 1001 <= logs.length <= 10^40 <= logs[i][0] <= 10^90 <= logs[i][1], logs[i][2] <= N - 1- It's guaranteed that all timestamps in logs[i][0] are different.
Logsare not necessarily ordered by some criteria.logs[i][1] != logs[i][2]
题解：
If log[1] and log[2] do NOT have same ancestor, put them into same union.
When count of unions become 1, that is the first time all people become friends and output time.
Time Complexity: O(mlogN). m = logs.length. find takes O(logN). With path compression and union by weight, amatorize O(1).
Space: O(N).
AC Java:
1 class Solution {
2 public int earliestAcq(int[][] logs, int N) {
3 UF uf= new UF(N);
4 Arrays.sort(logs, (a, b)-> a[0]-b[0]);
5
6 for(int [] log : logs){
7 if(uf.find(log[1]) != uf.find(log[2])){
8 uf.union(log[1], log[2]);
9 }
10
11 if(uf.count == 1){
12 return log[0];
13 }
14 }
15
16 return -1;
17 }
18 }
19
20 class UF{
21 int [] parent;
22 int [] size;
23 int count;
24
25 public UF(int n){
26 this.parent = new int[n];
27 this.size = new int[n];
28 for(int i = 0; i<n; i++){
29 parent[i] = i;
30 size[i] = 1;
31 }
32
33 this.count = n;
34 }
35
36 public int find(int i){
37 while(i != parent[i]){
38 parent[i] = parent[parent[i]];
39 i = parent[i];
40 }
41
42 return parent[i];
43 }
44
45 public void union(int p, int q){
46 int i = find(p);
47 int j = find(q);
48 if(size[i] > size[j]){
49 parent[j] = i;
50 size[i] += size[j];
51 }else{
52 parent[i] = j;
53 size[j] += size[i];
54 }
55
56 this.count--;
57 }
58 }
标签：Everyone,logs,int,timestamp,When,Moment,become,friends,log 来源： https://www.cnblogs.com/Dylan-Java-NYC/p/11280351.html
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