Find the number of reversed pairs in an array that does not have duplicate elements

problem:

Given a sequence a[n] with no repeated elements, ask how many pairs (i, j) exist, satisfying 0<=i<j<n && a[i] > a[j]

Solution: merge sort

The reverse logarithm of the sequence = the reverse logarithm of the left half + the reverse logarithm of the right half + the number of elements of the right half placed in front of the elements of the left half during the merge process

$$T(n)=2T(\frac{n}{2})+O\left(n\right)$$
so the time complexity is O(nlogn)

//合并a[0, n1)与a[n1, n)，返回合并中发现的逆序对数
int mergeArray(int a[], int n1, int n){
    
    
    int ans=0, p=0, q=n1;
    int *b = new int[n];
    memcpy(b, a, n*sizeof(int));
    for(int i=0; i<n; i++){
    
    
        //fill in a[i]
        if(q>=n || (p<n1 && b[p]<b[q])){
    
    
            //取前一半的数填，此时后半边已经填入了q-n1个，都和此数构成逆序对
            a[i] = b[p++];
            ans += q - n1;
        }
        else a[i] = b[q++];
    }
    delete[] b;
    return ans;
}

//返回a[0, n)中的逆序对数，同时把a归并排序
int mergeSort(int a[], int n){
    
    
    if(n==1) return 0;
    int mid = n/2;
    return mergeSort(a, mid) + mergeSort(a+mid, n-mid) + mergeArray(a, mid, n);
}