Screenshot from 2016-02-27 23:23:55.png

My code:

/**
 * 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;
        int[] starter = new int[intervals.length];
        int[] ender = new int[intervals.length];
        for (int i = 0; i < intervals.length; i++) {
            starter[i] = intervals[i].start;
            ender[i] = intervals[i].end;
        }=
        Arrays.sort(starter);
        Arrays.sort(ender);
        int endPoint = 0;
        int counter = 0;
        for (int i = 0; i < intervals.length; i++) {
            if (starter[i] < ender[endPoint]) {
                counter++;
            }
            else {
                endPoint++;
            }
        }
        return counter;
    }
}

这个方法比较巧，直接看的答案。
解释下，为什么 start[i] >= end[ep] 时，为什么endPointer++, i 也需要 ++
因为这个时候，start > end,
那么，时间上是独立的，不需要新开一间房。于是，start跟end都往后移动一格，这个时候再开始算，如果有重复的，再给room加上去。

参考网页：
https://leetcode.com/discuss/82292/explanation-super-easy-java-solution-beats-from-%40pinkfloyda

还有一个 O(n ^ 2)的做法，
参考网页是:
https://leetcode.com/discuss/86333/easy-and-concise-java-sol

Anyway, Good luck, Richardo!

My code:

/**
 * 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;
        }
        
        int[] start = new int[intervals.length];
        int[] end = new int[intervals.length];
        for (int i = 0; i < intervals.length; i++) {
            start[i] = intervals[i].start;
            end[i] = intervals[i].end;
        }
        Arrays.sort(start);
        Arrays.sort(end);
        
        int counter = 0;
        int endLocation = 0;
        for (int i = 0; i < start.length; i++) {
            if (start[i] < end[endLocation]) {
                counter++;
            }
            else {
                endLocation++;
            }
        }
        
        return counter;
    }
}

还是看了以前的答案做出来的。很巧妙。

先把start, end 都排序。
然后如果start >= end，那么说明，start这一个会议，可以和end那一组会议公用一个会议室。然后，这个会议室的结束时间就不再是现在的end了。而要往后延。
如果 start < end, 那么必须开一个会议室，counter++

Anyway, Good luck, Richardo! -- 09/15/2016