Problem statement:

Given an array of n integers where n > 1, nums, return an array output such that output[i] is equal to the product of all the elements of nums except nums[i].

Solve it without division and in O(n).

For example, given [1,2,3,4], return [24,12,8,6].

Follow up:
Could you solve it with constant space complexity? (Note: The output array does not count as extra space for the purpose of space complexity analysis.)

Solution one:

We can not do one traverse to get the total product of this array and traverse to get the product except for the element itself. If one element is 0 in this array, we can not get it return value.

We should analyze this problem the product except for the element itself is comprised of the element in front and at the back of it. The best solution does two directions traverse to find the product from left to it and the product form end to it.

The time complexity is O(n).

class Solution {public:    vector<int> productExceptSelf(vector<int>& nums) {        if(nums.empty()){            return {};        }        int frombegin = 1;        int fromlast = 1;        int size = nums.size();         vector<int> product(size, 1);                for(int i = 0; i < size; i++){            product[i] *= frombegin;            frombegin *= nums[i];            product[size - 1 - i] *= fromlast;            fromlast *= nums[size - 1 - i];        }        return product;    }};

238. Product of Array Except Self