Meeting room(sweep line)

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.

分析

按照起始时间排序，然后遍历起始每个Interval。判断起始时间是否晚于前面结束时间。

//  Definition for an interval.
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, new Comparator<Interval>() {
            @Override
            public int compare(Interval o1, Interval o2) {
                return o1.start - o2.start;
            }
        });

        for(int i = 1; i < intervals.length; i ++){
            if(intervals[i].start < intervals[i-1].end){
                return false;
            }
        }
        return true;
    }
}