Leetcode: 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.

F(0) = (0 * 4) + (1 * 3) + (2 * 2) + (3 * 6) = 0 + 3 + 4 + 18 = 25　　　　[4, 3, 2, 6]

F(1) = (0 * 6) + (1 * 4) + (2 * 3) + (3 * 2) = 0 + 4 + 6 + 6 = 16　　　　  [6, 4, 3, 2]        diff = (4+3+2) - 6*(nums.length-1)

F(2) = (0 * 2) + (1 * 6) + (2 * 4) + (3 * 3) = 0 + 6 + 8 + 9 = 23　　        [2, 6, 4, 3]　　  diff = (6+4+3) - 2*(nums.length-1)

 1 public class Solution {
 2     public int maxRotateFunction(int[] A) {
 3         if (A==null || A.length==0) return 0;
 4         int value = 0;
 5         int sum = 0;
 6         int maxValue = Integer.MIN_VALUE;
 7
 8         for (int i=0; i<A.length; i++) {
 9             value += i * A[i];
10             sum += A[i];
11         }
12
13         for (int j=A.length-1; j>=0; j--) {
14             value = value + (sum-A[j])-A[j]*(A.length-1);
15             maxValue = Math.max(maxValue, value);
16         }
17         return maxValue;
18     }
19 }