1288. Remove Covered Intervals (M)

Remove Covered Intervals (M)

题目

Given a list of intervals, remove all intervals that are covered by another interval in the list.

Interval [a,b) is covered by interval [c,d) if and only if c <= a and b <= d.

After doing so, return the number of remaining intervals.

Example 1:

Input: intervals = [[1,4],[3,6],[2,8]]
Output: 2
Explanation: Interval [3,6] is covered by [2,8], therefore it is removed.

Example 2:

Input: intervals = [[1,4],[2,3]]
Output: 1

Example 3:

Input: intervals = [[0,10],[5,12]]
Output: 2

Example 4:

Input: intervals = [[3,10],[4,10],[5,11]]
Output: 2

Example 5:

Input: intervals = [[1,2],[1,4],[3,4]]
Output: 1 

Constraints:

• 1 <= intervals.length <= 1000
• intervals[i].length == 2
• 0 <= intervals[i][0] < intervals[i][1] <= 10^5
• All the intervals are unique.

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题意

删除数组中所有被其他区间覆盖的区间。

思路

将数组按照左端点从小到大排序(相同则右端点大的在前)，从左到右遍历，记录右端点的最大值rightMax，如果当前区间的右端点小于rightMax，说明该区间被之前的某个区间覆盖，将其删去；否则更新rightMax。

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代码实现

Java

class Solution {
    public int removeCoveredIntervals(int[][] intervals) {
        int cnt = 0;
        Arrays.sort(intervals, (int[] a, int[] b) -> a[0] < b[0] ? -1 : a[0] == b[0] ? b[1] - a[1] : 1);
        int rightMax = intervals[0][1];
        for (int i = 1; i < intervals.length; i++) {
            if (intervals[i][1] <= rightMax) {
                cnt++;
            } else {
                rightMax = intervals[i][1];
            }
        }
        return intervals.length - cnt;
    }
}