Product of Array Except Self

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.)

public class Solution {
    public int[] productExceptSelf(int[] nums) 
    {
        int[] a = new int[nums.length];
        int[] b = new int[nums.length];
        int[] res = new int[nums.length];
        
        a[0] = 1;
        for(int i=1;i<nums.length;i++)
        {
            a[i] = a[i-1]*nums[i-1];
        }
        
        b[nums.length-1] = 1;
        for(int j=nums.length-2;j>=0;j--)
        {
            b[j] = b[j+1]*nums[j+1];
        }
        
        for(int s=0;s<nums.length;s++)
        {
            res[s] = a[s]*b[s];
            
        }
        return res;
    }
}

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.)

public class Solution {
    public int[] productExceptSelf(int[] nums) 
    {
        //first pass, {1,1,2,6}
        //second pass, {24,12,4,1}
        int[] res = new int[nums.length];
        res[0] = 1;
        for(int i=1; i< nums.length;i++)
        {
            res[i] = res[i-1]*nums[i-1];
        }
        int right = 1;
        
        for(int i=nums.length-1;i>=0;i--)
        {
            res[i] *= right;
            right *= nums[i];
        }
        return res;
    }
}

reference:https://leetcode.com/discuss/46104/simple-java-solution-in-o-n-without-extra-space