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:

1. You may assume the interval's end point is always bigger than its start point.
2. Intervals like [1,2] and [2,3] have borders "touching" but they don't overlap each other.

class Solution {
    public int eraseOverlapIntervals(int[][] intervals) {
        if(intervals.length == 0) return 0;
        Arrays.sort(intervals, (a, b) -> a[1] - b[1]);
        int res = 1;
        int curend = intervals[0][1];
        for(int i = 1; i < intervals.length; i++) {
            
 
if(curend <= intervals[i][0]) {
                res++;
                curend = intervals[i][1];  
            }            
        }
        return intervals.length - res;
    }
}

把問題轉化成：intervals中最多有多少個不重疊的array，然後用總數減去就是答案，和452.Minimum Number of Arrows to Burst Balloons，以及merge intervals挺像，前兩者都要先把陣列按end排序，這樣能為我們獲取更多的不重疊array