238. Product of Array Except Self（LeetCode）

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 sum = 1;
        for (int i = 0; i < nums.size(); i++)
        {
            sum *= nums[i];
        }
        if (sum == 0)
        {
            int count = 0;
            int flag = 0;
            for (int i = 0; i < nums.size(); i++)
            {
                if (nums[i] == 0)
                {
                    count++;
                    flag = i;
                }
            }
            if (count == 1)
            {
                int sum = 1;
                for (int i = 0; i < nums.size(); i++)
                {
                    if (i != flag)
                        sum *= nums[i];
                }
                for (int i = 0; i < nums.size(); i++)
                {
                    if (i == flag)
                        nums[i] = sum;
                    else
                        nums[i] = 0;
                }
            }
            else
            {
                for (int i = 0; i < nums.size(); i++)
                {
                    nums[i] = 0;
                }
            }
        }
        else
        {
            for (int i = 0; i < nums.size(); i++)
            {
                nums[i] = sum / nums[i];
            }
        }

        return nums;
    }
};

如果是考试，会立刻想到上面的方法，下面的方法是别人的，在考试的状态下，我感觉没有充裕的时间，我想不到。

class Solution {
public:
    vector<int> productExceptSelf(vector<int>& nums) {
        int n = nums.size();
        int fromBegin = 1;
        int fromLast = 1;
        vector<int> res(n, 1);

        for (int i = 0; i<n; i++){
            res[i] *= fromBegin;
            fromBegin *= nums[i];
            res[n - 1 - i] *= fromLast;
            fromLast *= nums[n - 1 - i];
        }
        return res;

    }
};