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.

Example

Given[[0, 30],[5, 10],[15, 20]], return false.

Solution

这题其实是考comparator的写法

1. 
Arrays.sort(intervals, new Comparator<Interval>() {
    public int compare(Interval a, Interval b) {
        return a.start - b.start;
    }
});

2.
Arrays.sort(intervals, new IntervalComparator());
private class IntervalComparator implements Comparator<Interval> {
    public int compare(Interval o1, Interval o2) {
    }
}

/**
 * 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 || intervals.length == 0) {
            return true;
        }

        Arrays.sort(intervals, new Comparator<Interval>() {
            public int compare(Interval a, Interval b) {
                return a.start - b.start;
            }
        });

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

        return true;
    }
}

改进一下compare函数，如果两个输入相交，直接throw exception

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();
    }
}