Rotate Function

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November 26, 2016 by oneOokay

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.

我觉得是一个数学题吧,需要观察找到规律,然后我想对啦 yayyy。

首先是brutal: 两个for-loop,TLE了;

接着稍微改变一下题目例子中的写法就可以找到规律:

F(0) = (0 * 4) + (1 * 3) + (2 * 2) + (3 * 6) = 25
F(1) = (1 * 4) + (2 * 3) + (3 * 2) + (0 * 6) = F(0) + (sum - 6) - (n - 1) * 6
F(2) = (2 * 4) + (3 * 3) + (0 * 2) + (1 * 6) = F(1) + (sum - 2) - (n - 1) * 2
F(3) = (3 * 4) + (0 * 3) + (1 * 2) + (2 * 6) = F(2) + (sum - 3) - (n - 1) * 3

F(n) = F(n- 1) + sum – A[n – 1] * n;

找到规律就好写了,其他没什么难度。就是要注意在Math.max的时候不要粗心把f[n]和f[n-1]对比。

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