435. Non-overlapping Intervals

Given an array of intervals intervals where intervals[i] = [starti, endi], return the minimum number of intervals you need to remove to make the rest of the intervals non-overlapping.

Example 1:

Input: intervals = [[1,2],[2,3],[3,4],[1,3]]
Output: 1
Explanation: [1,3] can be removed and the rest of the intervals are non-overlapping.

Example 2:

Input: intervals = [[1,2],[1,2],[1,2]]
Output: 2
Explanation: You need to remove two [1,2] to make the rest of the intervals non-overlapping.

Example 3:

Input: intervals = [[1,2],[2,3]]
Output: 0
Explanation: You don't need to remove any of the intervals since they're already non-overlapping.

Constraints:

• 1 <= intervals.length <= 2 * 104

• intervals[i].length == 2

• -2 * 104 <= starti < endi <= 2 * 104

Solution

有想到Sort但是sort要以end 當基底sort, 用start 去sort會有問題,假設有個interval的範圍很大但start很小

ex: [1,100000] 那之後interval都會因為這樣被刪除, 但理論上應該把這個interval刪掉就好了

class Solution {
    public int eraseOverlapIntervals(int[][] intervals) {
        Arrays.sort(intervals, (i, j)-> {
                int ret = i[1]-j[1]; 
                return ret;
            }
       );
        int count = 0, max = Integer.MIN_VALUE;
        for(int i = 0; i < intervals.length; i++){
            if(intervals[i][0] >= max){
                max = intervals[i][1];
                continue;
            }if(intervals[i][0] < max){
                count++;
            }
        }
        return count;
        
    }
}