class Solution {
public:
    vector<int> multiply(const vector<int>& A) {
        int n = A.size();
        if (!n) return A;                           //判空

        vector<int> B(n);

        for (int i = 0, p = 1; i <n; i ++ )         //先让B[i]等于i左半边的乘积
        {
            B[i] = p;
            p *= A[i];
        }

        for (int i = n - 1, p = 1; ~i; i -- )       //再用B[i]乘以i右半边的乘积
        {
            B[i] *= p;
            p *= A[i];
        }

        return B;
    }
};