问题：
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 = 25F(1) = (0 * 6) + (1 * 4) + (2 * 3) + (3 * 2) = 0 + 4 + 6 + 6 = 16F(2) = (0 * 2) + (1 * 6) + (2 * 4) + (3 * 3) = 0 + 6 + 8 + 9 = 23F(3) = (0 * 3) + (1 * 2) + (2 * 6) + (3 * 4) = 0 + 2 + 12 + 12 = 26So the maximum value of F(0), F(1), F(2), F(3) is F(3) = 26.
解决：
【题意】Bk是由A顺时针旋转k个位置得到的数组，求在旋转后的数组Bk中下标与对应数值的乘积的最大值。
① 暴力破解，O(N^2)，竟然没超时，但也没好到哪去。
class Solution {//640ms
public int maxRotateFunction(int[] A) { if(A == null || A.length == 0) return 0; int max = 0; int res = Integer.MIN_VALUE; for (int k = 0;k < A.length;k ++){ for (int i = 0;i < A.length;i ++){ max += i * A[i]; } res = Math.max(res,max); max = 0; int tmp = A[A.length - 1]; for (int j = A.length - 1;j > 0;j --){ A[j] = A[j - 1]; } A[0] = tmp; } return res; } }② 规律题
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：
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]
class Solution { //4ms
public int maxRotateFunction(int[] A) { if (A == null || A.length == 0) return 0; int len = A.length; int sum = 0; for (int i = 0;i < len;i ++){ sum += A[i]; } int f = 0; for (int i = 0;i < len;i ++){ f += i * A[i]; } int maxf = f; for (int k = 1;k < len;k ++){ f += (sum - len * A[len - k]); maxf = f > maxf ? f : maxf; } return maxf; } }