[Swift]LeetCode1101. 彼此熟识的最先时间 | The Earliest Moment When Everyone Become Friends

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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. 1 <= N <= 100
2. 1 <= logs.length <= 10^4
3. 0 <= logs[i][0] <= 10^9
4. 0 <= logs[i][1], logs[i][2] <= N - 1
5. It's guaranteed that all timestamps in logs[i][0] are different.
6. Logs are not necessarily ordered by some criteria.
7. logs[i][1] != logs[i][2]

---

在一个社交圈子当中，有 N 我的。每一个人都有一个从 0 到 N-1 惟一的 id 编号。

咱们有一份日志列表 logs，其中每条记录都包含一个非负整数的时间戳，以及分属两我的的不一样 id，logs[i] = [timestamp, id_A, id_B]。

每条日志标识出两我的成为好友的时间，友谊是相互的：若是 A 和 B 是好友，那么 B 和 A 也是好友。

若是 A 是 B 的好友，或者 A 是 B 的好友的好友，那么就能够认为 A 也与 B 熟识。

返回圈子里全部人之间都熟识的最先时间。若是找不到最先时间，就返回 -1 。

示例：

输入：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
输出：20190301
解释：
第一次结交发生在 timestamp = 20190101，0 和 1 成为好友，社交朋友圈以下 [0,1], [2], [3], [4], [5]。
第二次结交发生在 timestamp = 20190104，3 和 4 成为好友，社交朋友圈以下 [0,1], [2], [3,4], [5].
第三次结交发生在 timestamp = 20190107，2 和 3 成为好友，社交朋友圈以下 [0,1], [2,3,4], [5].
第四次结交发生在 timestamp = 20190211，1 和 5 成为好友，社交朋友圈以下 [0,1,5], [2,3,4].
第五次结交发生在 timestamp = 20190224，2 和 4 已是好友了。
第六次结交发生在 timestamp = 20190301，0 和 3 成为好友，你们都互相熟识了。

提示：

1. 1 <= N <= 100
2. 1 <= logs.length <= 10^4
3. 0 <= logs[i][0] <= 10^9
4. 0 <= logs[i][1], logs[i][2] <= N - 1
5. 保证 logs[i][0] 中的全部时间戳都不一样
6. Logs 不必定按某一标准排序
7. logs[i][1] != logs[i][2]

---

 368ms

 1 class Solution {
 2     func earliestAcq(_ logs: [[Int]], _ N: Int) -> Int {
 3         var logs = logs
 4         logs.sort(by:{return $0[0] < $1[0]})
 5         let dsu:DisjointSetUnion = DisjointSetUnion(N)
 6         for i in logs
 7         {
 8             dsu.AddEdge(i[1] + 1, i[2] + 1)
 9             if dsu.GetSize(1) == N
10             {
11                 return i[0]
12             }
13         }
14         return -1
15     }
16 }
17 
18 class DisjointSetUnion
19 {
20     var N_:Int
21     var par_:[Int]
22     var size_:[Int]
23     
24     init(_ N:Int)
25     {
26         self.N_ = N
27         self.par_ = [Int](repeating:0,count:N_ + 1)
28         self.size_ = [Int](repeating:1,count:N_ + 1)
29         for i in 0..<N
30         {
31             par_[i] = i
32         }
33     }
34     
35     // Returns the parent of |x|'s set.
36     func GetParent(_ x:Int) -> Int
37     {
38         if par_[x] != x
39         {
40             par_[x] = GetParent(par_[x])
41         }
42         return par_[x]
43     }
44     
45     // Returns the size of set in which |x| belongs.
46     func GetSize(_ x:Int) -> Int
47     {
48         return size_[GetParent(x)]
49     }
50     
51     // Add an edge between |u| and |v|. Creates union of both sets.
52     func AddEdge(_ u:Int,_ v:Int)
53     {
54         var parent_u:Int = GetParent(u)
55         var parent_v:Int = GetParent(v)
56         if parent_u != parent_v
57         {
58             if size_[parent_v] > size_[parent_u]
59             {
60                 let temp:Int = parent_u
61                 parent_u = parent_v
62                 parent_v = temp
63             }
64             par_[parent_v] = parent_u
65             size_[parent_u] += size_[parent_v]
66             size_[parent_v] = 0
67         }
68     }
69 }