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.

这道题一看，没啥思路，感觉应该有O(n)的解法，但是没想出来。

https://discuss.leetcode.com/topic/58616/java-solution-o-n-with-non-mathametical-explaination/3

Consider we have 5 coins A,B,C,D,E

According to the problem statement
F(0) = (0A) + (1B) + (2C) + (3D) + (4E)
F(1) = (4A) + (0B) + (1C) + (2D) + (3E)

F(2) = (3A) + (4B) + (0C) + (1D) + (2*E)

F(3) = (2A) + (3B) + (4C) + (0D) + (1*E)

F(4) = (1A) + (2B) + (3C) + (4D) + (0*E)

construct F(1) from F(0)

这是一道数学题，有个公式。看看连续状态之间的关系。

F(k) = 0 * Bk[0] + 1 * Bk[1] + ... + (n-1) * Bk[n-1]
F(k+1) = 0 * Bk+1[0] + 1 * Bk+1[1] + ... + (n-1) * Bk+1[n-1]
       = 0 * Bk[1] + 1 * Bk[2] + ... + (n-2) * Bk[n-1] + (n-1) * Bk[0]

Then,

F(k) - F(k+1) = Bk[1] + Bk[2] + ... + Bk[n-1] + (1-n)Bk[0]
              = (Bk[0] + ... + Bk[n-1]) - nBk[0]
              = sum - nBk[0]

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