252. Meeting Rooms

Given an array of meeting time intervals consisting of start and end times [[s1,e1],[s2,e2],...] (si < ei), determine if a person could attend all meetings.

For example,
Given [[0, 30],[5, 10],[15, 20]],
return false.

思路：把所有都interval按照start time从小到大sort一遍，然后挨个比较i和i+1的start和endtime使它们不相交，不重叠。如果重叠，就是false熟悉lamdba的用法

(a,b)->(a.val-b.val)  increasing order

(a,b) -> (b.val-a.val) decreasing order

/**
 * 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 boolean canAttendMeetings(Interval[] intervals) {
        Arrays.sort(intervals,(a,b)->(a.start-b.start));
        for(int i=0;i<intervals.length-1;i++)
        {
            if(intervals[i+1].start<intervals[i].end)
            {
                return false;
            }
        }
        return true;
    }
}