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.

求移除多少区间后，剩余区间都是不重叠的。

先求最多能组成多少不重叠的区间，再用总区间数减去不重叠区间数。

C++：

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