435. 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.

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.

 

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.

/**
 * Definition for an interval.
 * public class Interval {
 *     int start;
 *     int end;
 *     Interval() { start = 0; end = 0; }
 *     Interval(int s, int e) { start = s; end = e; }
 * }
 */
public class Solution {
    public int eraseOverlapIntervals(Interval[] intervals) {
        if(intervals.length == 0) return 0;
        Arrays.sort(intervals, new Comparator<Interval>(){
            public int compare(Interval a, Interval b){
                return a.start - b.start;
            }
        });
        int res = 0;
        int min = intervals[0].end;
        for(int i = 1 ; i < intervals.length ; i++){
            if(intervals[i].start < min){
                res++;
                min = Math.min(min, intervals[i].end);  // key point to make sure minimum : del the inteval with large end
            }
            else
              min = intervals[i].end;
        }
        return res;
    }
}

 

转载于:https://www.cnblogs.com/joannacode/p/6108943.html