238. Product of Array Except Self- 非自身數組的乘積

238. 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.)

public class Solution {
    public int[] productExceptSelf(int[] nums) {
        int count = 1;
        int count1 = 1;
        HashSet<Integer> set= new HashSet<Integer>();
        int[] result= new int[nums.length];
        for(int i = 0; i< nums.length; i++){
            count*=nums[i];
            if(nums[i]==0&&set.add(nums[i])) {
            	set.add(nums[i]);
            	continue;
            }
            count1*=nums[i];
        }
        for(int i = 0; i< nums.length; i++){
            if(nums[i]!=0) result[i]=count/nums[i];
            if(nums[i]==0) result[i]= count1;
        }
        return result;
    }
}

 感覺想的有點麻煩，應該有簡單方法。TO be  continue……

原文鏈接：https://leetcode.com/problems/product-of-array-except-self/