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.
Example 1:
Input:[[0,30],[5,10],[15,20]]
Output: false
Example 2:
Input: [[7,10],[2,4]]
Output: true
题目
给定一些区间，判断是否有重合。
还是挺实际的场景，经常在图书馆预定study room的系统，内部也应该是这个逻辑。 当有overlapping的时候，就会报错。
思路
任何一组intervals， 若当前start < 之前end，即出现了overlapping
代码
class Solution {
public boolean canAttendMeetings(Interval[] intervals) {
if(intervals == null || intervals.length==0) return true;
int []start = new int[intervals.length];
int []end = new int[intervals.length];
for(int i = 0; i<intervals.length;i++){
start[i] = intervals[i].start;
end[i] = intervals[i].end;
}
Arrays.sort(start);
Arrays.sort(end);
/* 任何一组intervals, 若当前start < 之前end，即出现了overlapping
| i-1 |
| i |
*/
for(int i = 1; i< start.length; i++){
if(start[i]<end[i-1]) return false;
}
return true;
}
}