Leetcode: 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.)

维持两个数组, left[] and right[]. left[i]记录第i个元素左边累乘的结果，right[i]表示第i个元素右边累乘的结果。那么结果res[i]即为  left[i]*right[i]. follow up 要求O(1)空间. 利用返回的结果数组,先存left数组. 再从右边计算right,同时计算结果值, 这样可以不需要额外的空间.

 1 public class Solution {
 2     public int[] productExceptSelf(int[] nums) {
 3         int[] res = new int[nums.length];
 4         res[0] = 1;
 5         for (int i=1; i<res.length; i++) {
 6             res[i] = res[i-1] * nums[i-1];
 7         }
 8         int right = 1;
 9         for (int i=res.length-1; i>=0; i--) {
10             res[i] = res[i] * right;
11             right = right * nums[i];
12         }
13         return res;
14     }
15 }