<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>

/**
 * 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* root = l2;
        int carry = 0;
        while (l2->next && l1->next) {
            l2->val = l2->val + l1->val + carry;
            carry = l2->val >= 10 ? 1 : 0;
            l2->val = l2->val % 10;
            l2 = l2->next;
            l1 = l1->next;
        }

        // Add final common node
        l2->val = l2->val + l1->val + carry;
        carry = l2->val >= 10 ? 1 : 0;
        l2->val = l2->val % 10;

        if (!l2->next) {
            l2->next = l1->next;
            // while (l1->next) {
            //     l2->next = new ListNode(l1->next->val + carry);
            //     l1 = l1->next;
            //     l2 = l2->next;
            //     if (l2->val >= 10) {
            //         carry = 1;
            //         l2->val -= 10;
            //     } else {
            //         carry = 0;
            //     }
            // }
        }


        // the rest is already here pretty much, just carry the final carry
        while (carry > 0) {
            if (l2->next) {
                l2 = l2->next;
                l2->val += carry;
                if (l2->val >= 10) {
                    carry = 1;
                    l2->val -= 10;
                } else {
                    carry = 0;
                }
            } else {
                l2->next = new ListNode(1);
                carry = 0;
            }
            
        }
        return root;
    }
};
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