<p>You are given two <strong>non-empty</strong> linked lists representing two non-negative integers. The digits are stored in <strong>reverse order</strong>, and each of their nodes contains a single digit. Add the two numbers and return the sum&nbsp;as a linked list.</p>

<p>You may assume the two numbers do not contain any leading zero, except the number 0 itself.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/10/02/addtwonumber1.jpg" style="width: 483px; height: 342px;" />
<pre>
<strong>Input:</strong> l1 = [2,4,3], l2 = [5,6,4]
<strong>Output:</strong> [7,0,8]
<strong>Explanation:</strong> 342 + 465 = 807.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> l1 = [0], l2 = [0]
<strong>Output:</strong> [0]
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> l1 = [9,9,9,9,9,9,9], l2 = [9,9,9,9]
<strong>Output:</strong> [8,9,9,9,0,0,0,1]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li>The number of nodes in each linked list is in the range <code>[1, 100]</code>.</li>
	<li><code>0 &lt;= Node.val &lt;= 9</code></li>
	<li>It is guaranteed that the list represents a number that does not have leading zeros.</li>
</ul>

#include<iostream>
using namespace std;

/**
 * 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) {}
 * };
 */

class Solution {
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        ListNode* dummyHead = new ListNode(0);
        ListNode* curr = dummyHead;
        int carry = 0;
        
        while (l1 != nullptr || l2 != nullptr || carry) {
            int x = (l1 != nullptr) ? l1->val : 0;
            int y = (l2 != nullptr) ? l2->val : 0;
            int sum = carry + x + y;
            
            carry = sum / 10;  // Calculate the carry for the next digit
            curr->next = new ListNode(sum % 10);  // Create a new node with the digit
            
            curr = curr->next;  // Move the current pointer to the next
            if (l1 != nullptr) l1 = l1->next;
            if (l2 != nullptr) l2 = l2->next;
        }
        
        return dummyHead->next;  // Return the actual result (skip the dummy node)
    }
};

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