Rotate Function

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Rotate Function

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:

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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.

Analysis

Java Solution O(n) with non mathametical explaination

Java O(n) solution with explanation_Math

一开始想排序后再求和，然后发现那样的话就变成乱序的了。

F(0) = (0A) + (1B) + (2C) + (3D) + (4E)

F(1) = (0B) + (1C) + (2D) + (3E) + (4A)

F(2) = (0C) + (1D) + (2E) + (3A) + (4B)

将数组逆序移动后总结规律可以发现F(n)等于F(n-1)减去所有元素的求和，再加上nums长度个上次贡献为0的元素，这样只需每次对tmp操作，并与res比较大小，取其大的即可。

Code

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public class Solution {
    public int maxRotateFunction(int[] A) {
        int res=0;
        int iteration=0;
        int size=A.length;
        for(int i=0;i<size;i++){
            res+=i*A[i];
            iteration+=A[i];
        }
        int tmp=res;
        for(int j=0;j<size-1;j++){
            tmp=tmp-iteration+size*A[j];
            res=Math.max(tmp,res);
        }
        return res;
    }
}