LeetCode 253. Meeting Rooms II（会议室）

原题网址：https://leetcode.com/problems/meeting-rooms-ii/

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.

方法一：对起始时间进行排序，使用最小堆来记录当前会议的结束时间，当心会议的起始时间大于最小堆中的最早结束时间，说明新会议与堆中的最早结束会议不重叠。

/**
 * 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) {
        Arrays.sort(intervals, new Comparator<Interval>() {
            @Override
            public int compare(Interval i1, Interval i2) {
                return Integer.compare(i1.start, i2.start);
            }
        });
        PriorityQueue<Integer> minHeap = new PriorityQueue<>();
        int rooms = 0;
        for(int i=0; i<intervals.length; i++) {
            minHeap.offer(intervals[i].end);
            if (intervals[i].start < minHeap.peek()) {
                rooms ++;
            } else {
                minHeap.poll();
            }
        }
        return rooms;
    }
}

更直观的实现方法：

/**
 * 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) {
        Arrays.sort(intervals, new Comparator<Interval>() {
            @Override
            public int compare(Interval i1, Interval i2) {
                return Integer.compare(i1.start, i2.start);
            }
        });
        int rooms = 0;
        int active = 0;
        PriorityQueue<Integer> heap = new PriorityQueue<>();
        for(int i=0; i<intervals.length; i++) {
            while (!heap.isEmpty() && heap.peek() <= intervals[i].start) {
                active --;
                heap.poll();
            }
            active ++;
            heap.offer(intervals[i].end);
            rooms = Math.max(rooms, active);
        }
        return rooms;
    }
}

方法二：分别对起始时间和结束时间排序，由于会议之间并无差异（不像skyline问题，不同建筑的高度不一样），所以分别使用两个指针来推进起始时间和结束时间。

/**
 * 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) {
        int[] starts = new int[intervals.length];
        int[] ends = new int[intervals.length];
        for(int i=0; i<intervals.length; i++) {
            starts[i] = intervals[i].start;
            ends[i] = intervals[i].end;
        }
        Arrays.sort(starts);
        Arrays.sort(ends);
        int rooms = 0;
        int activeMeetings = 0;
        int i=0, j=0;
        while (i < intervals.length && j < intervals.length) {
            if (starts[i] < ends[j]) {
                activeMeetings ++;
                i ++;
            } else {
                activeMeetings --;
                j ++;
            }
            rooms = Math.max(rooms, activeMeetings);
        }
        return rooms;
    }
}

参考文章：

http://buttercola.blogspot.com/2015/08/leetcode-meeting-rooms-ii.html

http://www.jyuan92.com/blog/leetcode-meeting-rooms-ii/