剑指offer 数组中的逆序对

class Solution {
public:
    long long InversePairsCore(vector<int> &data,vector<int>&copy,int start,int end)
    {
        if(start==end)
        {
            copy[start]=data[end];
            return 0;
        }
        int length=(end-start)/2;
        long long left = InversePairsCore(copy,data,start,start+length);
        long long right = InversePairsCore(copy,data,start+length+1,end);
        int i = start + length;
        int j = end;
        int indexCopy = end;
        long long  count=0;
        while(i>=start&&j>=start+length+1)
        {
            if(data[i]>data[j])
            {
                copy[indexCopy--]=data[i--];
                count+=j-start-length;
            }
             
            else
            {
                copy[indexCopy--]=data[j--];
            }
        }
        for(;i>=start;--i)
            copy[indexCopy--] = data[i];
        for(;j>=start+length+1;--j)
            copy[indexCopy--] = data[j];
        return left+right+count;
    }
    int InversePairs(vector<int> data) {
        int length=data.size();
        if(length<=0)
        {
            return 0;
        }
        vector<int> copy;
        for(int i=0;i<length;i++)
        {
            copy.push_back(data[i]);
        }
        long long count = InversePairsCore(data,copy,0,length-1);
        copy.clear();
        return count%1000000007;
    }
};