剑指offer第36题 求逆序数

#include<iostream>
using namespace std;

int Merge(int *list,int*copy,int s1,int s2,int end)
{
    int i=s1;
    int j=s2;
    int k=s1;
    int num=0;
    for(int k=s1;k<=end;k++)
        copy[k]=list[k];
    while(i<=s2-1&&j<=end)
    {
        if(copy[i]>copy[j])
        {
            num=num+s2-1-i+1;
            list[k]=copy[j];
            j++;
            k++;
        }
        else
        {
            list[k]=copy[i];
            i++;
            k++;
        }
    }
    while(i<=s2-1)
    {
        list[k]=copy[i];
        i++;
        k++;
    }
    while(j<=end)
    {
        list[k]=copy[j];
        j++;
        k++;
    }
    return num;
}

int InverseCore(int *list,int *copy,int start,int end)
{
    if(start==end)
        return 0;
    if(start<end)
    {
        int mid=(start+end)/2;
        int left=InverseCore(list,copy,start,mid);
        int right=InverseCore(list,copy,mid+1,end);
        int m=Merge(list,copy,start,mid+1,end);
        return left+right+m;
    }
}
int InversePairs(int *list,int length)
{
    if(list==NULL||length<0)
        return 0;
    int*copy=new int[length];//copy是辅助空间，作为参数传下去
    int count=InverseCore(list,copy,0,length-1);
    delete[]copy;
    copy=NULL;
    return count;
}

int main()
{
    int a[1000];
    int c;
    int i=0;
    while(cin>>c)
    {
        a[i]=c;
        i++;
        if(cin.get()==‘\n‘)
            break;
    }
    cout<<InversePairs(a,i);
}

使用递归的归并排序方法