Add Two Numbers II
<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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0400-0499/0445.Add%20Two%20Numbers%20II/images/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>
Stack; Linked List; Math
// Definition for singly-linked list.
// #[derive(PartialEq, Eq, Clone, Debug)]
// pub struct ListNode {
//   pub val: i32,
//   pub next: Option<Box<ListNode>>
// }
//
// impl ListNode {
//   #[inline]
//   fn new(val: i32) -> Self {
//     ListNode {
//       next: None,
//       val
//     }
//   }
// }
impl Solution {
    fn reverse(mut head: Option<Box<ListNode>>) -> Option<Box<ListNode>> {
        let mut pre = None;
        while let Some(mut node) = head {
            let next = node.next.take();
            node.next = pre.take();
            pre = Some(node);
            head = next;
        }
        pre
    }

    pub fn add_two_numbers(
        mut l1: Option<Box<ListNode>>,
        mut l2: Option<Box<ListNode>>,
    ) -> Option<Box<ListNode>> {
        l1 = Self::reverse(l1);
        l2 = Self::reverse(l2);
        let mut dummy = Some(Box::new(ListNode::new(0)));
        let mut cur = &mut dummy;
        let mut sum = 0;
        while l1.is_some() || l2.is_some() || sum != 0 {
            if let Some(node) = l1 {
                sum += node.val;
                l1 = node.next;
            }
            if let Some(node) = l2 {
                sum += node.val;
                l2 = node.next;
            }
            cur.as_mut().unwrap().next = Some(Box::new(ListNode::new(sum % 10)));
            cur = &mut cur.as_mut().unwrap().next;
            sum /= 10;
        }
        Self::reverse(dummy.unwrap().next.take())
    }
}
