LinkedList Questions: Add Two Numbers as Linked List
Tags: algorithms, codenewbie, programming, linkedlist
Published: 2021-06-26 04:28:29

In this series of posts, I will discuss coding questions on the `LinkedList` Data structure. 
The posts in this series will be organized in the following way,
1. Question Link ❓
2. Possible Explanation 📝
3. Documented C++ Code 🧹
4. Time and Space Complexity Analysis ⌛🌌

## The Question
You are given two **non-empty** linked lists representing two non-negative integers. The digits are stored in **reverse** order, and each of their nodes contains a single digit. Add the two numbers and return the sum as a linked list.

https://leetcode.com/problems/add-two-numbers/#

>💡 Give yourself atleast 15-20 mins to figure out the solution :)

## Explanation

Visualize how you used to do addition in your elementary school.

First create a dummynode whose `next` pointer will hold our resulting linkedlist. Make a `temp` pointer point to it. (it will be used for appending the resulting nodes)

>📒 The resulting linkedlist is also in reversed order

Then iterate through both the list, untill we reach the end in **both** the lists and there's no *carry* left.

- At every iteration, perform the arithmetic that we do while adding digits and calculate the resulting digit. 

- Create a newnode with value of resulting digit and append it to the end of our resulting linkedlist. (Notice the usecase of modulo operator).

## C++ Code

### Definition of LinkedList
```cpp 
//Definition for singly-linked list.
struct ListNode
{
    int val;
    ListNode *next;
    ListNode() : val(0), next(nullptr) {}
    ListNode(int x) : val(x), next(nullptr) {}
    ListNode(int x, ListNode *next) : val(x), next(next) {}
};
```

### Solution
```cpp
#include <bits/stdc++.h>
#include "../linkedlist.h"
using namespace std;

/*
-Time:O(max(n1, n2))
-Space:O(max(n1,n2))
*/
class Solution
{
public:
    //! Here we have to return the reversed list only
    ListNode *addTwoNumbers(ListNode *l1, ListNode *l2)
    {

        ListNode *start = new ListNode();
        ListNode *temp = start;

        //starting carry is zero
        int carry = 0;

        //go through both lists and create a new node untill
        //nodes exist in any of the lists or carry is 1
        while (l1 != nullptr || l2 != nullptr || carry != 0)
        {

            int sum = 0;

            if (l1 != nullptr)
            {
                sum += l1->val;
                l1 = l1->next;
            }

            if (l2 != nullptr)
            {
                sum += l2->val;
                l2 = l2->next;
            }

            sum = sum + carry;
            carry = sum / 10; //updating carry for next digit sum

            //note: We take modulo with 10
            temp->next = new ListNode(sum % 10);
            temp = temp->next;
        }

        return start->next;
    }
};

```
## Complexity Analysis
*n1* and *n2* are sizes of given linkedlists. 
### Time Complexity: O(max(n1, n2)) 
Since we have to travel both the lists completely.

### Space Complexity: O(max(n1, n2))
Same reason as above. 

>🔍Concretely both complexities will be **O(max(n1, n2) + 1)** by taking the end-carry into account but asymptotically, it doesn't matter.