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 = 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.

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这个题目的关键是我们不能每次都要全部乘一遍，我们算完第一次后，就可以利用第一次的结果，来计算下一个移动后的结果

计算公式为：temp = temp - (sum - A[i]) + A[i] * (A.size() - 1);temp为上一次的结果，sum为数组的和

class Solution {
public:
    int maxRotateFunction(vector<int>& A) {
        long long result = ;
        long long sum = ;
        long long temp = ;
        for (int i = ;i < A.size();++i)
        {
            result += A[i] * i;
            sum += A[i];
        }
        temp = result;
        for (int i = ;i < A.size();++i)
        {
            temp = temp - (sum - A[i]) + A[i] * (A.size() - );
            if (temp > result)
                result = temp;
        }
        return result;
    }
};