396. Rotate Function-蒲公英云

396. Rotate Function

Given an array of integers A and let n to be its length.

Assume Bk to be an array obtained by rotating the array A k positions clock-wise, we define a “rotation function” F on A as follow:

F(k) = 0 * Bk[0] + 1 * Bk[1] + ... + (n-1) * Bk[n-1].

Calculate the maximum value of F(0), F(1), ..., F(n-1).

Note:
n is guaranteed to be less than 105.

Example:

A = [4, 3, 2, 6]
F(0) = (0 * 4) + (1 * 3) + (2 * 2) + (3 * 6) = 0 + 3 + 4 + 18 = 25
F(1) = (0 * 6) + (1 * 4) + (2 * 3) + (3 * 2) = 0 + 4 + 6 + 6 = 16
F(2) = (0 * 2) + (1 * 6) + (2 * 4) + (3 * 3) = 0 + 6 + 8 + 9 = 23
F(3) = (0 * 3) + (1 * 2) + (2 * 6) + (3 * 4) = 0 + 2 + 12 + 12 = 26
So the maximum value of F(0), F(1), F(2), F(3) is F(3) = 26.
public class Solution {
    public int maxRotateFunction(int[] A) {
        if(A.length==0)
        return 0;
       int max = Integer.MIN_VALUE;
        for(int i=0;i<A.length;i++)
            if(F(A,i)>max)
                max = F(A,i);
        return max;
    }
    public int F(int[] A, int k) {
        int res = 0, count = 0;
        for (int i = A.length - k; i < A.length; i++)
            res += A[i] * count++;
        for (int i = 0; i < A.length - k; i++)
            res += A[i] * count++;
        return res;
    }
}