[Swift]LeetCode1101. 彼此熟识的最早时间 | 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 个人。每个人都有一个从 0 到 N-1 唯一的 id 编号。

我们有一份日志列表 logs，其中每条记录都包含一个非负整数的时间戳，以及分属两个人的不同 id，logs[i] = [timestamp, id_A, id_B]。

每条日志标识出两个人成为好友的时间，友谊是相互的：如果 A 和 B 是好友，那么 B 和 A 也是好友。

如果 A 是 B 的好友，或者 A 是 B 的好友的好友，那么就可以认为 A 也与 B 熟识。

返回圈子里所有人之间都熟识的最早时间。如果找不到最早时间，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，社交朋友圈如下 [0,1], [2], [3], [4], [5]。
第二次结交发生在 timestamp = 20190104，3 和 4 成为好友，社交朋友圈如下 [0,1], [2], [3,4], [5].
第三次结交发生在 timestamp = 20190107，2 和 3 成为好友，社交朋友圈如下 [0,1], [2,3,4], [5].
第四次结交发生在 timestamp = 20190211，1 和 5 成为好友，社交朋友圈如下 [0,1,5], [2,3,4].
第五次结交发生在 timestamp = 20190224，2 和 4 已经是好友了。
第六次结交发生在 timestamp = 20190301，0 和 3 成为好友，大家都互相熟识了。

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] 中的所有时间戳都不同
6. Logs 不一定按某一标准排序
7. logs[i][1] != logs[i][2]

---

 368ms

class Solution {
    func earliestAcq(_ logs: [[Int]], _ N: Int) -> Int {
        var logs = logs
        logs.sort(by:{return $0[0] < $1[0]})
        let dsu:DisjointSetUnion = DisjointSetUnion(N)
        for i in logs
        {
            dsu.AddEdge(i[1] + 1, i[2] + 1)
            if dsu.GetSize(1) == N
            {
                return i[0]
            }
        }
        return -1
    }
}

class DisjointSetUnion
{
    var N_:Int
    var par_:[Int]
    var size_:[Int]
    
    init(_ N:Int)
    {
        self.N_ = N
        self.par_ = [Int](repeating:0,count:N_ + 1)
        self.size_ = [Int](repeating:1,count:N_ + 1)
        for i in 0..<N
        {
            par_[i] = i
        }
    }
    
    // Returns the parent of |x|'s set.
    func GetParent(_ x:Int) -> Int
    {
        if par_[x] != x
        {
            par_[x] = GetParent(par_[x])
        }
        return par_[x]
    }
    
    // Returns the size of set in which |x| belongs.
    func GetSize(_ x:Int) -> Int
    {
        return size_[GetParent(x)]
    }
    
    // Add an edge between |u| and |v|. Creates union of both sets.
    func AddEdge(_ u:Int,_ v:Int)
    {
        var parent_u:Int = GetParent(u)
        var parent_v:Int = GetParent(v)
        if parent_u != parent_v
        {
            if size_[parent_v] > size_[parent_u]
            {
                let temp:Int = parent_u
                parent_u = parent_v
                parent_v = temp
            }
            par_[parent_v] = parent_u
            size_[parent_u] += size_[parent_v]
            size_[parent_v] = 0
        }
    }
}