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.)

思路：类似于dp的想法，先用left array存数字左边的所有乘积。然后再从右至左乘上右边的乘积。

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