构建乘积数组

链接：https://www.nowcoder.com/questionTerminal/94a4d381a68b47b7a8bed86f2975db46?answerType=1&f=discussion
来源：牛客网

既然不能用乘法，分析题目，我们可以将乘积拆为两项。即:

C[i] = A[0] * A[1] ...*A[i-1]
D[i] = A[i + 1] *... A[n-1]
B[i] = C[i] * D[i]
我们先来计算C[i]，使用数学归纳法：

C[0] = 1
C[1] = A[0]
C[2] = A[0] * A[1]
C[3] = A[0] * A[1] * A[2]
...
我们可以得出规律：C[i] = C[i-1] * A[i -1](i >=1)
我们继续用数学归纳法计算D[i]:

D[n - 1] = 1
D[n - 2] = A[n -1]
D[n - 3] = A[n - 1] * A[n - 2]
我们可以得出规律：D[i] = D[i + 1]* A[i + 1](i <= n-2)
代码：

class Solution {
public:
       //B[i] = C[i]*D[i];
       //C[i] = A[0]*A[1]*A[2]*A[i-1];
    //C[0] = 1,C[1] = A[0],C[2] = A[0]*A[1]
    //D[i] = A[i+1]*A[i+2]*A[n-1];
    //D[n-1] = 1,D[n-2] = A[n-1];
    vector<int> multiply(const vector<int>& A) {
        int n = A.size();
        vector<int> B(n,1);
        vector<int> C(n,1);
        vector<int> D(n,1);
        for(int i = 1;i<n;i++)
        {
            C[i] = C[i-1]*A[i-1];
        }

        for(int i = n-2;i>=0;i--)
        {
            D[i] = D[i+1]*A[i+1];
        }
        for(int i= 0;i<n;i++)
        {
            B[i] = C[i]*D[i];
        }
        return B;
    }
};