Product of Array Except Itself

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 ofnums 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.)

Analyse: Use a left vector and right vector to record the product until current number from left / right.  Space O(n), time O(n).

 1 class Solution {
 2 public:
 3     vector<int> productExceptSelf(vector<int>& nums) {
 4         vector<int> left(nums.size(), 1);
 5         vector<int> right(nums.size(), 1);
 6
 7         for(int i = 1; i < nums.size(); i++)
 8             left[i] = left[i - 1] * nums[i - 1];
 9         for(int i = nums.size() - 2; i >= 0; i--)
10             right[i] = right[i + 1] * nums[i + 1];
11
12         vector<int> result(nums.size(), 1);
13         for(int i = 0; i < nums.size(); i++)
14             result[i] = left[i] * right[i];
15
16         return result;
17     }
18 };

Use the result to store left production and a variable to store the right production. Space O(1), time O(n)

 1 class Solution {
 2 public:
 3     vector<int> productExceptSelf(vector<int>& nums) {
 4         int n = nums.size();
 5         vector<int> result(n, 1);
 6
 7         // first scan nums, and store its left production in result
 8         for(int i = 1; i < n; i++)
 9             result[i] = result[i - 1] * nums[i - 1];
10
11         int right = nums[n - 1];
12         for(int i = n - 2; i >= 0; i--) {
13             result[i] *= right;
14             right *= nums[i];
15         }
16         return result;
17     }
18 };