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.

沒什麼技巧，依次計算、比較即可

public class Solution {
    public static int maxRotateFunction(int[] A) {
        int n=A.length;
        int max=0;
        max=func(A,0);
        for(int i=1;i<A.length;i++){
            int temp=func(A,i);
            if(temp>max) max=temp;
        }
        return
 
 max;
    }
    public static int func(int[] A, int pos){
        int[] copy=new int[A.length];
        int p=0;
        for(int i=A.length - pos;i<A.length;i++){
            copy[p++]=A[i];
        }
        for(int i=0;i<A.length - pos;i++){
            copy[p++]=A[i];
        }
        int res=0;
        for(int i=0;i<copy.length;i++){
            res+=i*copy[i];
        }
        return res;
    }
}