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.)

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My draft solution:

public class Solution {    public int[] productExceptSelf(int[] nums) {        int[] output = new int[nums.length];output[0] = 1;for (int i = 1; i < nums.length; ++i) {output[i] = output[i - 1] * nums[i - 1];}int temp = nums[nums.length - 1];nums[nums.length - 1] = 1;for (int i = nums.length - 2; i >= 0; --i) {int n = nums[i];nums[i] = nums[i + 1] * temp;temp = n;}for (int i = 0; i < nums.length; ++i)output[i] *= nums[i];return output;    }}

And did some improvements:

public class Solution {    public int[] productExceptSelf(int[] nums) {        int[] output = new int[nums.length];output[0] = 1;for (int i = 1; i < nums.length; ++i) {output[i] = output[i - 1] * nums[i - 1];}for (int i = nums.length - 1,right = 1; i >= 0; --i) {output[i] *= right;right *= nums[i];}return output;    }}