（参考）Leetcode解题报告——396. Rotate Function

题目要求：
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).

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：
给定数组 A，设 0...n-1 与数组的乘积为 K ，依次颠倒数组 A ，求出最大 K 值

解题思路：
博主写了个O(n2)的方法，还是超时了，看到讨论区内都是通过找规律将时间复杂度减少到O(n)，哎，都怪数学不好。下面看一下别人的思路：

F(k) = 0 * Bk[0] + 1 * Bk[1] + ... + (n-1) * Bk[n-1]
F(k-1) = 0 * Bk-1[0] + 1 * Bk-1[1] + ... + (n-1) * Bk-1[n-1]
       = 0 * Bk[1] + 1 * Bk[2] + ... + (n-2) * Bk[n-1] + (n-1) * Bk[0]

F(k) - F(k-1) = Bk[1] + Bk[2] + ... + Bk[n-1] + (1-n)Bk[0]
              = (Bk[0] + ... + Bk[n-1]) - nBk[0]
              = sum - nBk[0]

k = 0; B[0] = A[0];
k = 1; B[0] = A[len-1];
k = 2; B[0] = A[len-2];

代码如下：

public  int maxRotateFunction(int[] A) {
        int sum = 0;
        int len = A.length;
        int F = 0;
        for (int i = 0; i < len; i++) {
            F += i * A[i];
            sum += A[i];
        }
        int max = F;
        for (int i = len - 1; i >= 1; i--) {
            F = F + sum - len * A[i];
            max = Math.max(F, max);
        }
        return max;
    }

参考资料：
Java O(n) solution with explanation