Description
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
.
Solution
Sort, time O(n * log n), space O(1)
比较排序后的intervals是否有重叠即可。
/**
* 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; }
* }
*/
class Solution {
public boolean canAttendMeetings(Interval[] intervals) {
if (intervals == null) {
return false;
}
Arrays.sort(intervals, new Comparator<Interval>(){
public int compare(Interval a, Interval b) {
return a.start != b.start ? a.start - b.start
: a.end - b.end;
}
});
for (int i = 1; i < intervals.length; ++i) {
if (intervals[i].start < intervals[i - 1].end) {
return false;
}
}
return true;
}
}