Leetcode 238 - 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.)
O(n), 所以只能扫一遍就得到该有的信息，自然想到的是左边扫一遍右边扫一遍。
省空间的话简单，右边扫的时候直接把答案存起来。

public class Solution {
    public int[] productExceptSelf(int[] nums) {
        //keep track of product of left elements in the array
        // then scan from right to left to multiply the right elements
        int n = nums.length;
        int[] results = new int[n];
        results[0] = 1;
        for (int i = 1; i < n; i++) {
            results[i] = results[i-1] * nums[i-1];
        }
        int right = 1;
        for (int i = n-1; i >= 0; i--) {
            results[i] *= right;
            right *= nums[i];
        }
        return results;
    }
}