396. Rotate Function

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题意： 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.

思路：这题如果用双层循环算最后是会报错的，所以肯定要找F的规律，既然是旋转函数不难看出:
F(1) = F(0)+sum(A)-(3*6)
F(2) = F(1)+sum(A)-(3*2)
F(3) = F(2)+sum(A)-(3*3)
F(4) = F(3)+sum(A)-(3*4)
即F(n) = F(n-1)+sum(A)-(len(A)*A[-1])，所以代码如下：

class Solution(object):    def maxRotateFunction(self, A):        """        :type A: List[int]        :rtype: int        """        f, s = 0, sum(A)        for i in xrange(len(A)):            f += i * A[i]        result = f        for i in xrange(1, len(A)+1):            f += s - len(A) * A[-i]            result = max(result, f)        return result

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