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 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.
给定一个数组Arr，设长度为len,遍历k∈[0,len)，计算
∑k*Arr[k]
然后把数组第一个元素放在最后面，这样执行len次。找到最大值。
使用数组大小两倍来存每次放在最后的元素，移动首指针，直接求解即可。
实现代码：
public class Solution {
public int MaxRotateFunction(int[] A) {
if(A == null || A.Length == 0){
return 0;
}
var len = A.Length;
var list = new int[2*len];
for (var i = 0;i < len; i++){
list[i] = A[i];
}
var max = int.MinValue;
var offset = 0;
for (var i = 0;i < len; i++){ // move number of 'len' numbers from first to last
var v = 0;
for (var j = offset;j < len+offset; j++){
v += list[j] * (j-offset);
}
if (max < v){
max = v;
}
// put first at last
var first = list[offset];
list[len + offset] = first;
// each move ,inc the offset
offset++;
}
//Console.WriteLine(list);
return max;
}
}