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.

这道题是LeetCode第四次比赛的第一道题，博主第一道题就没有做出来，博主写了个O(n²)的方法并不能通过OJ的大数据集合，后来网上看大家的解法都是很好的找到了规律，可以在O(n)时间内完成。现在想想找规律的能力真的挺重要，比如之前那道Elimination Game也靠找规律，而用傻方法肯定超时，然后博主发现自己脑子不够活，很难想到正确的方法，说出来全是泪啊T.T。好了，来解题吧，我们为了找规律，先把具体的数字抽象为A,B,C,D，那么我们可以得到：

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

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

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

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

那么，我们通过仔细观察，我们可以得出下面的规律：

sum = 1A + 1B + 1C + 1D

F(1) = F(0) + sum – 4D

F(2) = F(1) + sum – 4C

F(3) = F(2) + sum – 4B

那么我们就找到规律了, F(i) = F(i-1) + sum – n*A[n-i]，可以写出代码如下：

class Solution {
public:
    int maxRotateFunction(vector<int>& A) {
        int t = 0, sum = 0, n = A.size();
        for (int i = 0; i < n; ++i) {
            sum += A[i];
            t += i * A[i];
        }
        int res = t;
        for (int i = 1; i < n; ++i) {
            t = t + sum - n * A[n - i];
            res = max(res, t);
        }
        return res;
    }
};

参考资料：

https://leetcode.com/problems/rotate-function/

https://leetcode.com/problems/rotate-function/discuss/87853/Java-O(n)-solution-with-explanation

https://leetcode.com/problems/rotate-function/discuss/87842/Java-Solution-O(n)-with-non-mathametical-explaination