归并排序求逆序数

思路：假设前半部分（A[left]到A[mid]）与后半部分（A[mid+1]到A[right]）都是从小到大排好序的，那么如果A[left]>A[mid+1]，则A[mid+1]与前半部分的逆序数目是mid-left+1

int merge(int *A, int left, int mid, int right){
    int *temp = new int[right-left+1];
    int num = 0;
    int i = left;
    int j = mid+1;
    int index = 0;

    while(i<=mid && j<=right){
        if(A[i]>A[j]){
            num+=mid-i+1;
            temp[index]=A[j];
            j++;
        }
        else{
            temp[index]=A[i];
            i++;
        }
        index++;
    }
    if(i<=mid)
        for(;i<=mid;i++){
            temp[index]=A[i];
            index++;
        }
    if(j<=right)
        for(;j<=right;j++){
            temp[index]=A[j];
            index++;
        }

    for(int k=0; k<index; k++)
        A[left+k]=temp[k];
    
    delete []temp;
    return num;
}

int inversion(int *A, int left, int right){
    if(left>=right)  return 0;

    int mid=(left+right)/2;
    int num1=inversion(A,left,mid);
    int num2=inversion(A, mid+1, right);
    return num1+num2+merge(A,left,mid,right);
}

int main(){
    int A[5]={9,8,7,6,5};
    cout << inversion(A,0,4);
    return 0;
}