如何解决递归关系

Question

algorithm img is here!

sample(A[], p, r)
{
    if(p=r) return 1;
    sum = 0;
    for(int i = 1; i < n(?); i++)
        sum = sum + A[i];
    int q = (p+r)/2;
    int t = sum + sample(A,p,q) + sample(A, q+1, r);
    return t;
}

I use iteration method to solve this question in image. that code is right? i can't make recurrence relation. maybe T(n) = T...? where is "n"

Answer 1

You are breaking the size of recursion by factor 2, and also have a loop with n iteration. Hence, asymptotically, T(n) = 2T(n/2) + n = Theta(n log(n)) .