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.

此题是meeting room的一个变型题目，要求最少数量的重叠部分，那么就是相当于用贪心选择算法求出最多可容纳的interval的数量，然后来检验是否有和它重叠的，如果有重叠的部分，就coun++，思路和meeting room是一模一样的。代码如下：

/**

 * 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){

                if(a.end!=b.end){

                    return a.end-b.end;

                }else{

                    return a.start-b.start;

                }

            }

        });

        int end = intervals[0].end;

        int count = 0;

        for(int i=1;i<intervals.length;i++){

            if(intervals[i].start<end){

                count++;   

            }else{

                end = intervals[i].end;

            }

        }

        return count;

    }

}