252. Meeting Rooms

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.

按照会议开始时间排序，如果每个时间都是错开的，那么每一个interval[i].start会大于interval[i-1].end，否则就会有重叠。代码如下：

/**
 * 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; }
 * }
 */
public class Solution {
    public boolean canAttendMeetings(Interval[] intervals) {
        if (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);
        for (int i = 1; i < intervals.length; i ++) {
            if (start[i] < end[i - 1]) {
                return false;
            }
        }
        return true;
    }
}