*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.)

 

 
 1 public class Solution {
 2     public int[] productExceptSelf(int[] nums) 
 3     {
 4         int[] array1 = new int[nums.length];
 5         int[] array2 = new int[nums.length];
 6         int[] result = new int[nums.length];
 7 
 8         array1[0]=1;
 9         array2[nums.length-1]=1;
10         for(int i = 1; i<nums.length;i++)
11         {
12             array1[i]=array1[i-1]*nums[i-1];
13             
14         }
15         for(int i=nums.length-2;i>=0;i--)
16         {
17             array2[i]=array2[i+1]*nums[i+1];
18         }
19         for(int i=0;i<nums.length;i++)
20         {
21             result[i]=array1[i]*array2[i];
22         }
23         
24         return result;
25         
26         
27     }
28 }

 

posted @ 2015-10-08 04:48  Hygeia  阅读(128)  评论(0编辑  收藏  举报