<p>You are given two <strong>non-empty</strong> linked lists representing two non-negative integers. The most significant digit comes first and each of their nodes contains a single digit. Add the two numbers and return the sum 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/2021/04/09/sumii-linked-list.jpg" style="width: 523px; height: 342px;" />
<pre>
<strong>Input:</strong> l1 = [7,2,4,3], l2 = [5,6,4]
<strong>Output:</strong> [7,8,0,7]
</pre>

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

<pre>
<strong>Input:</strong> l1 = [2,4,3], l2 = [5,6,4]
<strong>Output:</strong> [8,0,7]
</pre>

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

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

<p>&nbsp;</p>
<p><strong>Follow up:</strong>&nbsp;Could you solve it without reversing the input lists?</p>

// 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) {
        vector<int> v1, v2;
        while (l1) {
            v1.push_back(l1->val);
            l1 = l1->next;
        }
        while (l2) {
            v2.push_back(l2->val);
            l2 = l2->next;
        }
        
        int carry = 0;
        ListNode* dummy = new ListNode(0);
        while (!v1.empty() || !v2.empty() || carry) {
            int sum = carry;
            if (!v1.empty()) {
                sum += v1.back();
                v1.pop_back();
            }
            if (!v2.empty()) {
                sum += v2.back();
                v2.pop_back();
            }
            carry = sum / 10;
            ListNode* node = new ListNode(sum % 10);
            node->next = dummy->next;
            dummy->next = node;
        }
        
        return dummy->next;
    }
};
78000