Rotate Function

Given an array of integersAand letnto be its length.

AssumeBkto be an array obtained by rotating the arrayAkpositions clock-wise, we define a "rotation function"FonAas follow:

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

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

Note:
nis 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.

Tips:

列式子做一下比较，就会发现

F(i + 1) = F(i) + sum - A.length * A[i]

Code:

public class Solution {
    public int maxRotateFunction(int[] A) {
        int sum = 0;
        int F = 0;
        for (int i = 0; i < A.length; i++) {
            F += i * A[i];
            sum += A[i];
        }
        int max = F;
        for (int i = A.length - 1; i >= 0; i--) {
            F = F + sum - A[i] * A.length;
            max = Math.max(F, max);
        }
        return max;
    }
}