problem

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 = 25F(1) = (0 * 6) + (1 * 4) + (2 * 3) + (3 * 2) = 0 + 4 + 6 + 6 = 16F(2) = (0 * 2) + (1 * 6) + (2 * 4) + (3 * 3) = 0 + 6 + 8 + 9 = 23F(3) = (0 * 3) + (1 * 2) + (2 * 6) + (3 * 4) = 0 + 2 + 12 + 12 = 26So the maximum value of F(0), F(1), F(2), F(3) is F(3) = 26.

solution

先遍历下标，然后遍历元素，性能没过关；嵌套次反过来呢？,

怎么嵌套大致都是O(n^2^)，差别不大

class Solution(object):    def maxRotateFunction(self, A):        """        :type A: List[int]        :rtype: int        """        sums = []        length = len(A)        for i in range(length):            sum = 0            for index,item in enumerate(A):                sum += (index+i)%length * item            sums.append(sum)        if  not len(sums):            sums.append(0)        return max(sums)

class Solution(object):    def maxRotateFunction(self, A):        """        :type A: List[int]        :rtype: int        """        length = len(A)        sums = [0]*length                for index,item in enumerate(A):            for offset in range(length):                sums[offset] += (index+offset)%length * item        if  not len(sums):            sums.append(0)        return max(sums)

discuss

数学寻找关联，然后求解

F(k) = 0 * Bk[0] + 1 * Bk[1] + ... + (n-1) * Bk[n-1]F(k-1) = 0 * Bk-1[0] + 1 * Bk-1[1] + ... + (n-1) * Bk-1[n-1]       = 0 * Bk[1] + 1 * Bk[2] + ... + (n-2) * Bk[n-1] + (n-1) * Bk[0]Then,F(k) - F(k-1) = Bk[1] + Bk[2] + ... + Bk[n-1] + (1-n)Bk[0]              = (Bk[0] + ... + Bk[n-1]) - nBk[0]              = sum - nBk[0]Thus,F(k) = F(k-1) + sum - nBk[0]What is Bk[0]?k = 0; B[0] = A[0];k = 1; B[0] = A[len-1];k = 2; B[0] = A[len-2];...

int maxRotateFunction(int* A, int ASize) {    int i, last, sum,m;        if(ASize<=1) return 0;        last=0;    sum=0;    for(i=0;i
    
     =1;i--){        last += sum - (ASize*A[i]);        if (last > m) m= last;    }        return m;}