Merge Intervals

Given a collection of intervals, merge all overlapping intervals.

For example,
Given [1,3],[2,6],[8,10],[15,18],
return [1,6],[8,10],[15,18].

/* *
 * 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 {
private:
     void quicksort(vector<Interval> & datas, int left, int right)
    {
         if(left>=right) 
             return;
         int l=left;
         int r=right;
         int m=datas[(l+r)/ 2].start;
         while(l<=r)
        {
             while(datas[l].start<m) l++;
             while(datas[r].start>m) r--;
             if(l>right || r<left || l>=r)
                 break;
            
            Interval tmp=datas[l];
            datas[l]=datas[r];
            datas[r]=tmp;
            l++;r--;
        }
        
        quicksort(datas,left,l- 1);
        quicksort(datas,r+ 1,right);
    }
    inline  bool checkoverlap( const Interval& interval1, const Interval& interval2)
    {
         if(interval1.start<=interval2.start && interval1.end>=interval2.end)  return  true;
         if(interval1.start<=interval2.start && interval1.end>=interval2.start)  return  true;
         if(interval1.start<=interval2.end && interval1.end>=interval2.end)  return  true;
         return  false;
    }
    inline  void mergeinterval(Interval& interval1, const Interval& interval2)
    {
         if(interval1.start>interval2.start) interval1.start=interval2.start;
         if(interval1.end<interval2.end) interval1.end=interval2.end;
    }
public:
    vector<Interval> merge(vector<Interval> &intervals) 
    {
        vector<Interval> v;
         if(intervals.size()== 0)  return v;
        
         // sort by start
        quicksort(intervals, 0,intervals.size()- 1);
         // merge
        
         int index= 0;
        v.push_back(intervals[ 0]);
        
         for( int i= 1;i<intervals.size();i++)
        {
             if(checkoverlap(v[index],intervals[i]))
            {
                mergeinterval(v[index],intervals[i]);
            }
             else
            {
                v.push_back(intervals[i]);
                index++;
            }
        }
        
         return v;
    }
};