Question
Given an array of meeting time intervals consisting of start and end times [[s1,e1],[s2,e2],...]
(si < ei), find the minimum number of conference rooms required.
For example,
Given [[0, 30],[5, 10],[15, 20]]
,
return 2
.
Difficulty:Medium
Category:
Solution
先把所有的时间区间按照起始时间排序,然后新建一个最小堆,开始遍历时间区间,如果堆不为空,且首元素小于等于当前区间的起始时间,我们去掉堆中的首元素,把当前区间的结束时间压入堆,由于最小堆是小的在前面,那么假如首元素小于等于起始时间,说明上一个会议已经结束,可以用该会议室开始下一个会议了,所以不用分配新的会议室,遍历完成后堆中元素的个数即为需要的会议室的个数
Solution 1: Priority_queue
Time complexity: O(n log n), Space complexity: O(n)
class Solution {
public:
int minMeetingRooms(vector<Interval> &intervals) {
sort(intervals.begin(), intervals.end(), [](const Interval &a, const Interval &b) { return a.start < b.start; });
priority_queue<int, vector<int>, greater<int>> q;
for (auto a : intervals) {
if (!q.empty() && q.top() <= a.start) q.pop();
q.push(a.end);
}
return q.size();
}
};