leetcode435. Non-overlapping Intervals

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.

    public int eraseOverlapIntervals(int[][] intervals) {
        if(intervals == null || intervals.length == 0) return 0;
        Arrays.sort(intervals, new Comparator<int[]>() {

            @Override
            public int compare(int[] o1, int[] o2) {
                // TODO Auto-generated method stub
                return o1[0] - o2[0] == 0 ? o1[1] - o2[1] : o1[0] - o2[0];
            }
            
        });
        
        int count = 0;
        int end = intervals[0][1];
        for(int i = 1 ; i<intervals.length ; i++) {
            int[] interval = intervals[i];
            if(interval[0] < end) {;
                count++;
                end = Math.min(end, interval[1]);
            }else {
                end = interval[1];
            }
        }
        return count;
    }