leetcode253

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.
Example 1:
Input: [[0, 30],[5, 10],[15, 20]]
Output: 2
Example 2:
Input: [[7,10],[2,4]]
Output: 1

最小堆。O(nLogn), O(n)
预处理：先把interval按照interval的start从小到大排好序。开一个最小堆，使得堆顶是end最小的interval。
遍历intervals，把当前interval的start和堆顶的end对比，如果交叉，说明不能用最早结束的会议室从而要重开一个新房间，入当前，heap.size()多了1说明多一个会议室；如果不交叉，说明可以之前那个房间开会的人走了我可以直接用那个房间了，出堆顶入当前，heap.size()不变说明不用多会议室。
返回的答案也就是这个过程中heap.size()的巅峰值，打擂台得到。

实现

/**
 * Definition for an interval.
 * public class Interval {
 *     int start;
 *     int end;
 *     Interval() { start = 0; end = 0; }
 *     Interval(int s, int e) { start = s; end = e; }
 * }
 */
class Solution {
    public int minMeetingRooms(Interval[] intervals) {
        if (intervals == null || intervals.length == 0) {
            return 0;
        }
        Arrays.sort(intervals, new Comparator<Interval>() {
            @Override
            public int compare(Interval a, Interval b) {
                return a.start - b.start;
            }
        });
        
        PriorityQueue<Interval> minHeap = new PriorityQueue<>(new Comparator<Interval>() {
            @Override
            public int compare(Interval a, Interval b) {
                return a.end - b.end;
            }
        });
        
        int ans = 1;
        minHeap.offer(intervals[0]);
        for (int i = 1; i < intervals.length; i++) {
            int earliestEnd = minHeap.peek().end;
            if (intervals[i].start >= earliestEnd) {
                minHeap.poll();
                minHeap.offer(intervals[i]);
            } else {
                minHeap.offer(intervals[i]);
            }
            ans = Math.max(ans, minHeap.size());
        }
        return ans;
        
    }
}