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.

思路: 参考了某位博主的blog,规律如下
F(i) = F(i-1) + sum – n*A[n-i]

public class Solution {
    public int maxRotateFunction(int[] A) {
        int f = 0, n = A.length, sum = 0;

        for (int i = 0; i < n; i++) { // calculate F[0] and sum
            sum += A[i];
            f += i* A[i];
        }

        int res = f; // assign F(0)
        for (int i = 1; i < n; i++) {
            f = f + sum - n*A[n-i];
            res = Math.max(res, f);
        }
        return res;
    }
}

Reference:
396. Rotate Function

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