Interview Question 36 reverse order in the array

The title describes the two numbers in the array, and if the previous number is greater than the number that follows, the two numbers form an inverse pair. Enter an array to find the total number of reverse pairs in this array.

1 classSolution {2  Public:3     intInversepairscore (vector<int> &data,vector<int>&copy,intStartintend)4     {5         if(start==end)6         {7copy[start]=Data[start];8             return 0;9         }Ten         intLength = (end-start)/2; One         intleft = Inversepairscore (copy,data,start,start+length); A         intright = Inversepairscore (copy,data,start+length+1, end); -            -         inti = start +length; the         intj =end; -         intIndexcopy =end; -         intCount=0; -          while(i>=start&&j>=start+length+1) +         { -             if(data[i]>Data[j]) +             { Acopy[indexcopy--]=data[i--]; atcount+=j-start-length; -             } -              -             Else -             { -copy[indexcopy--]=data[j--]; in             } -         } to          for(; i>=start;--i) +copy[indexcopy--] =Data[i]; -          for(; j>=start+length+1;--j) thecopy[indexcopy--] =Data[j]; *         returnleft+right+count; $         }Panax Notoginseng     intInversepairs (vector<int>data) { -         intLength =data.size (); the         if(length<=0) +             return 0; Avector<int>copy; the          for(intI=0; i<length;i++) + Copy.push_back (Data[i]); -         intCount = Inversepairscore (Data,copy,0, length-1); $ copy.clear (); $         returncount; -     } -       the       -};

Interview Question 36 reverse order in the array