模板题目：Interval (LeetCode253：Meeting Rooms I: I, LeetCode729: My Calendar I)

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.

这道题可以说是greedy的经典题目了。我们把这道题复述一下，其实意思就是求所有时刻的最大深度，返回这个深度就是返回最少用几个房间。

class Solution {
    public int minMeetingRooms(int[][] intervals) {
        if(intervals == null || intervals.length == 0) return 0;
        
        int[] starts = new int[intervals.length];
        int[] ends = new int[intervals.length];
        
        for (int i = 0; i<intervals.length; i++) {
            starts[i] = intervals[i][0];
            ends[i] = intervals[i][1];
        }
        Arrays.sort(starts);
        Arrays.sort(ends);
        
        int end = 0;
        int res = 0;
        
        for (int i = 0; i < intervals.length; i++) {
            if (starts[i] < ends[end]) {
                res++; //still don't understand why res keeps culmulate even after we change the pointer end???我现在也不明白
            } else {
                end++;
            }
        }
        return res;
    }
}

我也不知道这种题目有啥模板可言 如果非要说有的话 那就是Greedy吧