LeetCode 396. Rotate Function

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.

刚开始用暴力枚举，时间复杂度O(n^2)，而n<=10^5，果断超时。

没有思路了，然后参考了大神的解法。

如果向右移动一位，那么除了最后A[n-1]，每个数都多加了一次，即多加了A[0]+A[1]+...+A[n-2]，即sum-A[n-1];A[n-1]变成0,即减少了(n-1)*A[n-1];

所以，f(n+1)=f(n)+A[0]+A[1]+...+A[n-2]-(n-1)*A[n-1]=f(n)+sum-A[n-1]-(n-1)*A[n-1]=f(n)+n*A[n-1].

int maxRotateFunction(int* A, int ASize) {    if(!ASize) return 0;    int sum = 0, i;    int f = 0, max;    max = 1 << 31;    for(i = 0; i < ASize; i ++){        sum += A[i];        f += i * A[i];    }    for(i = 0; i < ASize; i ++){        f = f + sum - ASize * A[ASize - i - 1];        if(f > max) max = f;    }    return max;}