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.

一刷
先根据start时间排序，再判断有没有overlap

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

二刷
同上

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