Rotate Function问题及解法

问题描述：

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 105.

示例:

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)
4000
+ (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.

问题分析：

假设我们数组中有5个值a,b,c,d,e，根据问题描述我么可以求得如下式子

F(0) = (0a) + (1b)
+ (2c) + (3d) + (4e)

F(1) = (4a) + (0b) + (1c)
+ (2d) + (3e)
F(2) = (3a) + (4b)
+ (0c) + (1d) + (2e) 

F(3) = (2a) + (3b)
+ (4c) + (0d) + (1e)
F(4) = (1a) + (2b)
+ (3c) + (4d) + (0e) 

我们发现由F(k)到F(k + 1)的关系

F(k+1) = F(k)  - sum + A[k] * len,sum是数组的和，len是数组的长度。

过程详见代码：

class Solution {
public:
int maxRotateFunction(vector<int>& A) {
int n = A.size();
if (!n) return 0;
long res = INT_MIN;
long f = 0;
for (int k = 0; k < n; k++)
{
f = 0;
for (int j = 0; j < n; j++)
{
f += j * A[(n - k + j) % n];
}
if (f > res) res = f;
}
return res;
}
};