Product of Array Exclude Itself

Given an integers array A.

Define B[i] = A[0] * ... * A[i-1] * A[i+1] * ... * A[n-1], calculate B WITHOUT divide operation.

Have you met this question in a real interview? 

Yes

Example

For A = [1, 2, 3], return [6, 3, 2].

按照定義做，定義兩個數組A,B，A[i]表示i位置後的數的乘積，B[i]表示i前的數的乘積，這樣ans[i] = A[I] * B[i];

class Solution {
public:
    /**
     * @param A: Given an integers array A
     * @return: A long long array B and B[i]= A[0] * ... * A[i-1] * A[i+1] * ... * A[n-1]
     */
    //分段的思路，用兩個數組B[i]，表示i之前的數的乘積， A[i]表示i之後的數的乘積
    vector<long long> productExcludeItself(vector<int> &nums) {
        // write your code here
        int len = nums.size();
        vector<long long> ans;
        if(len < 1)
            return ans;
        long long  *A = new long long[len + 5];
        long long *B = new long long[len + 5];
        for(int i = 0; i < len + 5; ++i)
        {
            A[i] = 1;
            B[i] = 1;
        }
        
        for(int i = 1; i < len; ++i)
        {
            B[i] = nums[i - 1] * B[i - 1];
        }
        
        for(int i = len - 2; i >= 0; --i)
        {
            A[i] = A[i + 1] * nums[i + 1];
        }
        
        for(int i = 0; i < len; ++i)
        {
            ans.push_back(A[i] * B[i]);
        }
        
        return ans;
    }
};