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 = 25F(1) = (0 * 6) + (1 * 4) + (2 * 3) + (3 * 2) = 0 + 4 + 6 + 6 = 16F(2) = (0 * 2) + (1 * 6) + (2 * 4) + (3 * 3) = 0 + 6 + 8 + 9 = 23F(3) = (0 * 3) + (1 * 2) + (2 * 6) + (3 * 4) = 0 + 2 + 12 + 12 = 26So the maximum value of F(0), F(1), F(2), F(3) is F(3) = 26.

关键在于如何进行转换。

public class Solution {

    public int maxRotateFunction(int[] A) {
        
        int n=A.length;
        if(n==0)
        return 0;
        int max=0;
        int s=0,res=0;
       for(int i=0;i<n;i++){
        s=s+A[i];
        max=max+i*A[i];
        }
        res=max;
      for(int i=1;i<n;i++){
     res=res+s-A[n-i]*n;
     if(res>max)
     max=res;
      }
        return max;
    }
}