题目：
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.
链接： http://leetcode.com/problems/meeting-rooms/
题解：
一开始以为是跟Course Schedule一样，仔细读完题目以后发现只要sort一下就可以了。写得还不够精简，需要好好研究一下Java8的lambda表达式。
Time Complexity - O(nlogn)， Space Complexity - O(1)
/** * 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) { if(intervals == null || intervals.length == 0) return true; Arrays.sort(intervals, new Comparator<Interval>(){ public int compare(Interval t1, Interval t2) { if(t1.start != t2.start) return t1.start - t2.start; else return t1.end - t2.end; } }); for(int i = 1; i < intervals.length; i++) { if(intervals[i].start < intervals[i - 1].end) return false; } return true; } }
Reference:
https://leetcode.com/discuss/50912/ac-clean-java-solution
原文：http://www.cnblogs.com/yrbbest/p/5012310.html