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.)

public class Solution {    public int[] productExceptSelf(int[] nums) {        int[] result = new int[nums.length];        if(nums == null || nums.length == 0){            return result;        }        result[result.length-1] = 1;//把最右边的数设为1        for(int i= result.length-2; i>=0; i--){            result[i] = result[i+1]*nums[i+1];        }//从右往左扫一遍，逐步相乘，这样到最左边的时候就是除了nums[0]和nums[nums.length-1]以外所有数的乘积        int leftSum = 1;        for(int i=0; i<nums.length; i++){            result[i] *= leftSum;            leftSum *= nums[i]; //再从左往右扫一遍即可        }        return result;            }}