<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>

static const int fastIO = [] {
	std::ios_base::sync_with_stdio(false), std::cin.tie(nullptr), std::cout.tie(nullptr);
	return 0;
}();

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;
		}
		std::reverse(v1.begin(), v1.end());
		std::reverse(v2.begin(), v2.end());
		int i = 0, j = 0, carry = 0;
		vector<int> ans;
		while (i < v1.size() && j < v2.size()) {
			int sum = v1[i++] + v2[j++] + carry;
			ans.push_back(sum % 10);
			carry = sum / 10;
		}
		while (i < v1.size()) {
			int sum = v1[i++] + carry;
			ans.push_back(sum % 10);
			carry = sum / 10;
		}
		while (j < v2.size()) {
			int sum = +v2[j++] + carry;
			ans.push_back(sum % 10);
			carry = sum / 10;
		}
		if (carry) ans.push_back(carry);
		ListNode* head = new ListNode(ans.front());
		for (int k = 1; k < ans.size(); k++)
			head = new ListNode(ans[k], head);
		return head;
	}
};
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