Given an array of meeting time intervals intervals where intervals[i] = [starti, endi], return the minimum number of conference rooms required.
Solution
我们用一个小根堆（优先队列）来保存结束时间，每次比较当前的开始时间和队首元素，如果不重叠，则可以 \(pop\) 出去，代表用了当前的房间，不必添加新的房间
点击查看代码
class Solution {
private:
int ans=0;
priority_queue<int,vector<int>,greater<int>> no_use;
public:
int minMeetingRooms(vector<vector<int>>& intervals) {
sort(intervals.begin(), intervals.end());
int n = intervals.size();
no_use.push(intervals[0][1]);
for(int i=1;i<n;i++){
if(no_use.top()<=intervals[i][0])no_use.pop();
no_use.push(intervals[i][1]);
}
return no_use.size();
}
};