Rotate Function解题报告

Description:

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).

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.

Link:

https://leetcode.com/problems/rotate-function/#/description

解题方法：

1、暴力破解，通过F(k)的方程对每次进行环位移的数组来求出解，也能AC。
2、通过观察得知，F(k)都可以由F(k-1)求出，方程为F(i-1) + sum - n*A[n-i] = F(i)，其中sum为数组元素的和。所以求出F(0)和sum就可以把剩下的算出。

Time Complexity：

解法一：O(N^2)
解法二：O(N)

完整代码：

解法一：
int maxRotateFunction(vector& A) 
    {
        int max = INT_MIN;
        if(A.size() == 0)
            return 0;
        for(int k = 0; k < A.size(); k++)
        {
            int sum = 0;
            int i = 0;
            for(; i < k; i++)
                sum += (A[A.size() - k + i] * i);
            for(int j = 0; j < A.size() - k; j++, i++)
                sum += (A[j] * i);
            max = sum > max ? sum : max;
        }
        return max;
    }
解法二：
int maxRotateFunction(vector& A) 
    {
        if(A.size() == 0)
            return 0;
        int f = 0, sum = 0, n = A.size();
        for(int i = 0; i < n; i++)
        {
            f += A[i] * i;             //算出F(0)
            sum += A[i];               //算出sum
        }
        int max = f;
        for(int i = 1; i < n; i++)
        {
            f = f + sum - (n * A[n - i]);   //F(i-1) + sum - n*A[n-i] = F(i)             
            max = f > max ? f : max;    
        }
        return max;
    }