# 396. Rotate Function 旋转函数

## # 题目描述：

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 10^5.

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.
``````

## # 解题方法

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

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

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

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

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

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

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

``````class Solution:
def maxRotateFunction(self, A):
"""
:type A: List[int]
:rtype: int
"""
_sum = 0
N = len(A)
f = 0
for i, a in enumerate(A):
_sum += a
f += i * a
res = f
for i in range(N - 1, 0, -1):
f = f + _sum - N * A[i]
res = max(res, f)
return res
``````

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

## # 日期

2018 年 10 月 10 日 ———— 冻成狗