Meeting Rooms2

题目地址：
https://leetcode.com/problems/meeting-rooms-ii/#/description

题目：

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.

解题思路：
这道题首先要根据所有interval的start来排序，先把第一个interval放进minHeap，minHeap是根据interval的end来排序。然后循环interval array，先poll出一个interval，判断该interval的end和当前interval的start的关系，如果start小于interval的end，那么必须要多加一个教室；反之则更新之前poll出来的interval。

代码：

public int minMeetingRooms(Interval[] intervals) {
    if(intervals == null || intervals.length == 0){
        return 0;
    }
    Arrays.sort(intervals, new Comparator<Interval>(){
        public int compare(Interval a, Interval b){
            return a.start - b.start;
        }
    });
    PriorityQueue<Interval> minHeap = new PriorityQueue<Interval>(intervals.length, new Comparator<Interval>(){
        public int compare(Interval a, Interval b){
            return a.end - b.end;
        }
    });
    minHeap.offer(intervals[0]);
    for(int i = 1; i < intervals.length; i++){
        Interval interval = minHeap.poll();
        if(intervals[i].start >= interval.end){
            interval.end = intervals[i].end;
        }
        else{
            minHeap.offer(intervals[i]);
        }
        minHeap.offer(interval);
    }
    return minHeap.size();
}