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.

此题重点是理解题意，会议时间不可以存在交集。

其次，了解排序，即数组有数组排序，Arrays.sort. Collections.sort. PriorityQueue()三种排序方式，都可以重写来实现。

代码如下：

/**
 * 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;
        Comparator<Interval> comp = new Comparator<Interval>(){
            public int compare(Interval i1,Interval i2){
                return i1.start-i2.start;
            }
        };
        Arrays.sort(intervals,comp);
        //int start  = intervals[0].start;
        int end = intervals[0].end;
        for(int i=1;i<intervals.length;i++){
            if(intervals[i].start-end<0) return false;
            else{
                end = Math.max(end,intervals[i].end);
            }
        }
        return true;
    }
}
//the run time could include the time sorting and the compare, so the total run time could be O(nlongn), the sace complexity could be O(1);