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.

O(nlogn).

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public class Solution {
	public boolean canAttendMeetings(Interval[] intervals) {
	    // Sort the intervals by start time
	    Arrays.sort(intervals, (x, y) -> x.start - y.start);
	    for (int i = 1; i < intervals.length; i++)
	        if (intervals[i-1].end > intervals[i].start)
	            return false;
	    return true;
	}
}

The following solution is cited from Leetcode Discussion board.

####Find an overlap while sorting.

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private boolean canAttendMeetings(Interval[] intervals) {
    try {
        Arrays.sort(intervals, new IntervalComparator());
    } catch (Exception e) {
        return false;
    }
    return true;
}

private class IntervalComparator implements Comparator<Interval> {
    @Override
    public int compare(Interval o1, Interval o2) {
        if (o1.start < o2.start && o1.end <= o2.start)
            return -1;
        else if (o1.start > o2.start && o1.start >= o2.end)
            return 1;
        throw new RuntimeException();
    }
}