396. Rotate Function

Problem:

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.

Analysis:

We can convert each number to A,B,C,D, then we get

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

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

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

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

From there we can conclude that

F(1) = F(0) + sum - 4D

F(2) = F(1) + sum - 4C

F(3) = F(2) + sum - 4B

then F(i) = F(i-1) + sum - n*A[n-i]

http://www.cnblogs.com/grandyang/p/5869791.html

Solution:

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