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.)

---

解题思路：

参考答案。

Because we cannot use division, so assume we have two integer arrays with the same length of nums, 

int[] leftProd = new int[nums.length]; int[] rightProd = new int[nums.length],

we store the product of all the left elements in leftProd and the product of all the right elements in rightProd,

then the product of leftProd[i] and rightProd[i] will be the value we want to put into the result.

Take the example of num[] = {2, 4, 3, 6}, then leftProd will be {1, 2, 8, 24} , and rightProd will be {72, 18, 6, 1}.

---

Java code:

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

---

Reference:

1. https://leetcode.com/discuss/46150/java-o-n-solution-no-extra-space-with-explanation