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).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.Bk代表对数组A在位置k上进行顺时针的旋转后生成的数组。F(k) = 0 * Bk[0] + 1 * Bk[1] + … + (n-1) * Bk[n-1]，要求返回获得的最大的F(k)的值。暴力循环按照题目的要求，执行两次循环即可以获得F(k)的所有值，只需要从中比较最大值即可。 public int maxRotateFunction(int[] A) { if(A == null || A.length == 0) return 0; int max = Integer.MIN_VALUE; for(int i = 0 ; i < A.length ; i++) { int value = 0; for(int j = 0 ; i < A.length ; j++) { value += j * A[(j+i)%A.length]; } max = Math.max(value, max); } return max; }数学思路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] F(k) = F(k-1) + sum - n*Bk[0]k = 0 Bk[0] = A[0]k = 1 Bk[0] = A[len-1]k = 2 Bk[0] = A[len-2]… public int maxRotateFunction(int[] A) { if(A == null || A.length == 0) return 0; int F = 0; int sum = 0; for(int i = 0 ; i<A.length ; i++) { sum += A[i]; F += i * A[i]; } int max = F; for(int i = 1 ; i<A.length ; i++) { F += sum - A.length * A[A.length - i]; max = Math.max(F, max); } return max; }