Rotate Function 396

Question

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.

Solution

这道题如果直接计算每一次的值的话一定会超时，因此必须寻找每次值之间的关系规律。

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：

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]

代码如下：

public class Solution {
    public int maxRotateFunction(int[] A) {
        if (A == null || A.length <= 1) {
            return 0;
        }

        //F(i) = F(i- 1) + sum - n * A[n - i]
        int n = A.length;
        int sum = 0;
        int pre = 0;
        for (int i = 0; i < n; i++) {
            sum += A[i];
            pre += i * A[i];
        }

        System.out.println(sum);
        System.out.println(pre);

        int max = pre;
        for (int i = 1; i < n; i++) {
            int curt = pre + sum - n * A[n - i];
            max = Math.max(max, curt);
            pre = curt;
        }

        return max;
    }
}