<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 {
	pair<ListNode*, int> solve(ListNode* l1, int n, ListNode* l2, int m) {
		if (n == 0 || m == 0) return { nullptr, 0};
		if (n > m) {
			auto ans = solve(l1->next, n - 1, l2, m);
			int sum = l1->val + ans.second;
			return { new ListNode(sum % 10, ans.first), sum / 10 };
		}
		if (n < m) {
			auto ans = solve(l1, n, l2->next, m - 1);
			int sum = l2->val + ans.second;
			return { new ListNode(sum % 10, ans.first), sum / 10 };
		}
		auto ans = solve(l1->next, n - 1, l2->next, m - 1);
		int sum = l1->val + l2->val + ans.second;;
		return { new ListNode(sum % 10, ans.first), sum / 10 };
	}

public:
	ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
		int n = 0, m = 0;
		auto aux = l1;
		while (aux) {
			n++;
			aux = aux->next;
		}
		aux = l2;
		while (aux) {
			m++;
			aux = aux->next;
		}
		auto ans = solve(l1, n, l2, m);
		if (ans.second)
			return new ListNode(ans.second, ans.first);
		return ans.first;
	}
};
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