# LeetCode 396. Rotate Function

## Description

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.
```

## 描述

F(k) = 0 * Bk[0] + 1 * Bk[1] + … + (n-1) * Bk[n-1]。

```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

```

### 思路

• 移动原数组和移动索引效果相同（第一次原数组乘[0,1..n-1]，第二次原数组乘[n-1,0,1…]）
• 观察发现：第 i+1 次的结果可以由第 i 次结果得到。如下图：
```# -*- coding: utf-8 -*-
# @Author:             何睿
# @Create Date:        2019-08-28 21:06:46

from typing import List

class Solution:
def maxRotateFunction(self, A: List[int]) -> int:
sum_, count = sum(A), len(A)
product = sum(map(lambda x: x[0] * x[1], enumerate(A)))
res = product

for i in A[:-1]:
# 上一次的乘积减去所有数的和，加上当前位置的数乘以总数的个数
product = product - sum_ + count * i
res = max(res, product)

return res
```