【LeetCode】1101. The Earliest Moment When Everyone Become Friends 解题报告 (C++)

作者：负雪明烛
id： fuxuemingzhu
个人博客：http://fuxuemingzhu.cn/

题目描述

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:

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

题目大意

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 。

解题方法

并查集

提示的不能更明显了，标准的并查集。

对logs按照时间排序。
遍历logs，合并两个人所属的环，如果环减少到1那就是最短的时间。

C++代码如下：

class Solution {
public:
    int earliestAcq(vector<vector<int>>& logs, int N) {
        map_ = vector<int>(N);
        circle = N;
        for (int i = 0; i < N; ++i)
            map_[i] = i;
        sort(logs.begin(), logs.end(), [](vector<int>& a, vector<int>& b) {return a[0] < b[0];});
        for (auto& log : logs) {
            uni(log[1], log[2]);
            if (circle == 1)
                return log[0];
        }
        return -1;
    }
    int find(int a) {
        if (map_[a] == a)
            return a;
        return find(map_[a]);
    }
    void uni(int a, int b) {
        int pa = find(a);
        int pb = find(b);
        if (pa == pb)
            return;
        map_[pa] = pb;
        circle --;
    }
private:
    vector<int> map_;
    int circle = 0;
};

日期

2019 年 9 月 21 日 —— 莫生气，我若气病谁如意