[Leetcode] Product of Array Except Self, Solution

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.)

[Thoughts]

通常這種題均可以分解成若干子問題來解決。As defined, 

output[i] is equal to the product of all the elements of  nums except  nums[i].

簡單的說

    output[i] =  { i 前面的數的乘積}  X  { i 後面的數的乘積}

問題就解決了，首先從前日後掃描數組一遍，對每個i，獲得{i 前面的數的乘積}(能夠稱作output_before)，而後在從後往前掃描數組一遍，得到 { i 後面的數的乘積}(能夠稱作output_after)。 將兩數相乘即爲所求。

舉個例子(以下圖)，nums = {1,2,3,4}, 第一遍，從前日後掃描一遍，獲得的output_before = {1, 1, 2, 6}. 從後往前掃描一遍，獲得的output_after = {24, 12, 4, 1}.

那麼  output [i] = output_before[i] * output_after[i],   output = {24, 12, 8, 6}

[Code]

1:  class Solution {  
2:  public:  
3:    vector<int> productExceptSelf(vector<int>& nums) {  
4:      vector<int> products;  
5:      if(nums.size() == 0) return products;  
6:      int product = 1;  
7:      products.push_back(1);  
8:      for(int i =1; i< nums.size(); i++) {  
9:        product *= nums[i-1];  
10:        products.push_back(product);  
11:      }  
12:      product = 1;  
13:      for(int i =nums.size()-2; i>=0; i--) {  
14:        product = product * nums[i+1];  
15:        products[i] = products[i] * product;   
16:      }  
17:      return products;  
18:    }  
19:  };  

github:  https://github.com/codingtmd/leetcode/blob/master/src/Product%20of%20Array%20Except%20Self.cpp