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.

思路：貪心演算法，以每個元素的end為關鍵元素排序，end相等的情況下，按start升序排序。這裡的實際型別和這個問題很像，

只有一個場地，要安排儘量多的活動。貪心規則：儘量安排結束時間早的活動。如果後面的活動與已經安排好的相容，則加入集合。

/**
 * Definition for an interval.
 * struct Interval {
 *     int start;
 *     int end;
 *     Interval() : start(0), end(0) {}
 *     Interval(int s, int e) : start(s), end(e) {}
 * };
 */
class Solution {
    
public:
    int eraseOverlapIntervals(vector<Interval>& intervals) {
       int n=intervals.size();
       if(!n) return 0;
       sort(intervals.begin(),intervals.end(),[](Interval a,Interval b){
           return a.end<b.end||(a.start>b.start&&a.end==b.end);
       });
       int cnt=1, s=intervals[0].start,e=intervals[0].end;
       for(int i=1;i<n;i++)
       {
           //if(it.start==s&&it.end>e) continue;
           Interval it=intervals[i];
           if(it.start>=e) {
               cnt++;
              // s=it.start;
               e=it.end;
           }
       }
       return n-cnt;
    }
};