396 Rotate Function

396 Rotate Function

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.

分析& 代码

　　s0 = 0 * a0 + 1 *a1 + 2 * a2 + 3 * a3

-　s1 = 1 * a0 + 2 * a1 + 3 * a2 + 0 * a3

-------------------------------------------------

　- a0  - a1 - a2 + 3*a3 = - (a0+a1+a2+a3)+4*a4

所以如果sum表示所有数的和，len表示数组的长度那么每连续两个的差别是：

s0 - s1 = -sum + len*a3

所以公式就是

sn = sn-1 + sum - len * a[len - n]

所以代码是：

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