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class Solution {
    public boolean canAttendMeetings(int[][] intervals) {
        Arrays.sort(intervals, (a, b) -> Integer.compare(a[0], b[0]));
        for (int i = 1; i < intervals.length; i++) {
            if (intervals[i][0] < intervals[i - 1][1])
                return false;
        }
        return true;
    }
}

Merge intervals的简单版本.

时间复杂度: O(nlogn)
空间复杂度: O(logn)
排序消耗时间.