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.)

class Solution {public:    vector<int> productExceptSelf(vector<int>& nums)     {        int n = nums.size();        vector<int> result(n, 0);        if(n == 0)            return result;        int all = 1;     //nums中所有数字的乘积        int cnt = 0;     //nums中0的个数        int index = -1;  //nums中哪个数字是0        for(int i=0; i<n; i++)        {            if(nums[i] != 0)            {                all *= nums[i];            }            else            {                cnt++;                index = i;                if(cnt == 2)                    return result;             }        }        if(cnt == 2) //nums中有2个0        {            return result;        }        else if(cnt == 1)  //nums中有1个0        {            result[index] = all;            return result;        }        //正常情况        for(int i=0; i<n; i++)        {            if(i != index)                result[i] = all / nums[i];        }        return result;    }};