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.)

此题解法是进行两次遍历，第一次从前往后遍历，第二次从后往前遍历，没遍历到一个数组值的时候，计算该数组值前遍历过的值的乘积，然后保存在乘积的数组里面，代码如下：

public class Solution {

    public int[] productExceptSelf(int[] nums) {

        int[] res=  new int[nums.length];

        int left = 1;

        int right = 1;

        res[0] = 1;

        for(int i=1;i<nums.length;i++){

            res[i] = res[i-1]*nums[i-1];

        }

        for(int i=nums.length-1;i>=0;i--){

            res[i] = res[i]*right;

            right = right*nums[i];

        }

        return res;

    }

}

238. Product of Array Except Self