396. Rotate Function (python)

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.
题意：将数组A元素右移K位得到数组B，然后计算 函数F(0), F(1), …, F(n-1)的最大值
方法一：每次将数组右移一位然后进行函数运算，时间复杂度为0（n^2），空间复杂度为O（n）
运行会出现超时
方法二：从F(0)和F(1)的关系可以看出，F(1)-F(0)为全部元素之和减去最后一个元素*n
因此，可推导：F(i) = F(i-1) + sum(A) -n*A[n-i]
这样时间复杂度降为O(n)
但需要注意全部元素之和只计算一次，不要在循环中多次计算，不然也会超时
Runtime: 69 ms

class Solution(object):    def rotate(self,nums,k):        n=len(nums)        if n<2 :            return nums        k=k%n        nums=nums[n-k:]+nums[:n-k]        return nums    def maxRotateFunction(self, A):        if len(A)==0:            return 0        f=[]        for i in range(len(A)):            B=self.rotate(A,i)            s=0            for j in range(len(B)):                s+=B[j]*j            f.append(s)        return max(f)

class Solution(object):    def maxRotateFunction(self, A):        n=len(A)        if n==0:            return 0        s=0        sA=0        for i in range(n):            s+=A[i]*i            sA+=A[i]        maxf=s        for i in range(1,n):            s=s+sA-n*A[n-i]            maxf=max(maxf,s)        return maxf