参考：https://leetcode-cn.com/problems/maximum-number-of-events-that-can-be-attended/solution/sao-miao-suan-fa-tan-xin-by-lucifer1004/

class Solution {
public:
    int maxEvents(vector<vector<int>>& events) {
        sort(events.begin(), events.end()); //以开始时间排序升序
        int res = 0; 
        priority_queue<int, vector<int>, greater<int> >pq;  //小根堆
        for(int i = 1,j = 0; i < 100005; i++)
        {//1 <= events[i][0] <= events[i][1] <= 10^5
            for( ; j < events.size() && events[j][0] <= i; j++)
            {//将开始时间早于等于时间i的会议的结束时间存入小根堆
             //不是每次都会进这个循环
                pq.push(events[j][1]);
            }
            while(!pq.empty())
            {
                int end = pq.top();
                pq.pop();
                if(end >= i ) 
                {//如果i > end则会议结束时间比i早,会议过期了
                    res++; 
                    break;
                }
            }
        }
        return res;
    }
};