## DEV Community

Kathan Vakharia

Posted on • Updated on

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,

2. Possible Explanation π
3. Documented C++ Code π§Ή
4. Time and Space Complexity Analysis βπ

## 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.

π‘ Give yourself atleast 15-20 mins to figure out the solution :)

## Explanation

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)

π 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 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

``````#include <bits/stdc++.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 *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.

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