排序,然后比较开始结束时间即可
时间O(nlogn)(主要消耗在排序上),空间O(1)
public boolean canAttendMeetings(int[][] intervals) { Arrays.sort(intervals, new Comparator<int[]>() { public int compare(int[] i1, int[] i2) { return i1[0] - i2[0]; } }); for (int i = 0; i < intervals.length - 1; i++) { if (intervals[i][1] > intervals[i + 1][0]) return false; } return true; }