Given a collection of intervals, find the minimum number of intervals you need to remove to make the rest of the intervals non-overlapping.
Example 1:
Input: [[1,2],[2,3],[3,4],[1,3]]
Output: 1
Explanation: [1,3] can be removed and the rest of intervals are non-overlapping.
Example 2:
Input: [[1,2],[1,2],[1,2]]
Output: 2
Explanation: You need to remove two [1,2] to make the rest of intervals non-overlapping.
Example 3:
Input: [[1,2],[2,3]]
Output: 0
Explanation: You don't need to remove any of the intervals since they're already non-overlapping.
Note:
You may assume the interval's end point is always bigger than its start point.
Intervals like [1,2] and [2,3] have borders "touching" but they don't overlap each other.
classSolution {publicinteraseOverlapIntervals(int[][] intervals) {if (intervals.length==0)return0;Arrays.sort(intervals, (a, b) -> a[1] - b[1]);int end = intervals[0][1];int count =1;// Count total number of non overlapping intervalsfor (int i =1; i <intervals.length; i++) {if (intervals[i][0] >= end) { end = intervals[i][1]; count++; } }returnintervals.length- count; }}