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.)

C++ 代码如下（涵盖测试代码）：

#include <iostream>
#include <vector>

using namespace std;

class Solution {
public:
    vector<int> productExceptSelf(vector<int>& nums) {
        if(nums.size() <= 0)
                return nums;
        int products = 1;
        vector<int> zeroNum;
        for(int i = 0; i < nums.size(); i++)
                if(nums[i] != 0)
                        products *=  nums[i];
                else {
                        zeroNum.push_back(i);
                }

        //cout << "output" << endl; 
        vector<int> result(nums.size(), 0);
        if(zeroNum.size() >= 2) {
                return result;
        }

        if(zeroNum.size() == 1) {
                result[zeroNum[0]] = products;
                return result;
        }
        for(int i = 0; i < nums.size(); i++) {
                result[i] = products/nums[i] ;
        }
        return result;
    }
};

int main() {
        vector<int> nums;
        nums.push_back(0);
        nums.push_back(2);
        nums.push_back(0);
        nums.push_back(4);

        Solution s;
        vector<int> r = s.productExceptSelf(nums);
        for(int i = 0; i < r.size(); i++)
                cout << r[i] << endl;
}

注：提交两遍才过，第一遍忘记了考虑整个数组中有 0 的情况，加上判断，就通过了。

思路很简单，关键是细节。