396. Rotate Function - jiejackyzhang/leetcode-note GitHub Wiki

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.

设数组A长度为n，所有数之和为sum。 这道题的规律为：

F(1) = F(0) + sum - A[n-1]*n

F(2) = F(1) + sum - A[n-2]*n

...

F(i) = F(i-1) + sum - A[n-i]*n for 1 <= i < n

public class Solution {
    public int maxRotateFunction(int[] A) {
        if(A == null || A.length == 0) return 0;
        int sum = 0, f = 0;
        int n = A.length;
        for(int i = 0; i < n; i++) {
            sum += A[i];
            f += A[i] * i;
        }
        int res = f;
        for(int i = n-1; i > 0; i--) {
            f = f + sum - A[i] * n;
            if(f > res) res = f;
        }
        return res;
    }
}