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