## 题目描述：

LeetCode 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.
```

## 题目大意：

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

n确保小于10^5

## 解题思路：

```0 1 2 3 4
1 2 3 4 0
2 3 4 0 1
3 4 0 1 2
4 0 1 2 3```

## Python代码：

``````class Solution(object):
def maxRotateFunction(self, A):
"""
:type A: List[int]
:rtype: int
"""
size = len(A)
sums = sum(A)
sumn = sum(x * n for x, n in enumerate(A))
ans = sumn
for x in range(size - 1, 0, -1):
sumn += sums - size * A[x]
ans = max(ans, sumn)
return ans
``````

Pingbacks已关闭。

1. Joe Hisaishi 发布于 2016年9月15日 03:01 #

-- 用每一行系数与其上一行做差，差值恰好为<code>sum(A) + size * A[size - x]</code>
++ 用每一行系数与其上一行做差，差值恰好为<code>sum(A) - size * A[size - x]</code>

2. 在线疯狂 发布于 2016年9月15日 09:50 #

感谢更正，已经修改 :D