LeetCode 253. Meeting Rooms II

原题链接在这里：https://leetcode.com/problems/meeting-rooms-ii/

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.

题解：

排序后扫描，同时维护一个min heap, 把每个interval 的end 放到min heap中. 若是新的interval start 大于 heap peek, 就一直heap poll.

maxOverlap是这个过程中heap size的峰值。

Time Complexity: O(nlogn). Space: O(n).

AC Java:

/**
 * 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; }
 * }
 */
public class Solution {
    public int minMeetingRooms(Interval[] intervals) {
        if(intervals == null || intervals.length == 0){
            return 0;
        }
        
        Arrays.sort(intervals, new Comparator<Interval>(){
            public int compare(Interval i1, Interval i2){
                if(i1.start == i2.start){
                    return i1.end - i2.end;
                }
                return i1.start - i2.start;
            }
        });
        
        int maxOverlap = 0;
        PriorityQueue<Integer> minHeap = new PriorityQueue<Integer>();
        for(int i = 0; i<intervals.length; i++){
            minHeap.add(intervals[i].end);
            while(!minHeap.isEmpty() && minHeap.peek()<=intervals[i].start){
                minHeap.poll();
            }
            maxOverlap = Math.max(maxOverlap, minHeap.size());
        }
        return maxOverlap;
    }
}

类似Car Pooling, Meeting Rooms.