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.

思路

利用最简单的方法会超时。

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class Solution(object):
    def maxRotateFunction(self, A):
        """
        :type A: List[int]
        :rtype: int
        """
        length  = len(A)
        res = []
        for i in range(length):
            tmp = 0
            for j in range(length):
                tmp += j * A[j-i]
            res.append(tmp)
        if (len(res)):
            return max(res)
        else:
            return 0

把函数变换一下。得到一个关系式，如：

F(1) = F(0) + A数组的和 - (元素个数 * 最后一个元素)

F(2) = F(1) + A数组的和 - (元素个数 * 倒数第二个元素)

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class Solution(object):
    def maxRotateFunction(self, A):
        """
        :type A: List[int]
        :rtype: int
        """
        length  = len(A)
        if (length):
            suma = 0
            for i in A:
                suma += i
            res = []
            cur = 0
            for i in range(length):
                cur += A[i] * i
            res.append(cur)
            
            for i in range(length-1):
                prev = res[-1]
                cur = prev + suma - (A[-i-1] * length)
                res.append(cur)
            
            return max(res)
        else:
            return 0