Function leetcode_for_rust::cd0002_add_two_numbers::add_two_numbers

pub fn add_two_numbers(
    l1: Option<Box<ListNode>>, 
    l2: Option<Box<ListNode>>
) -> Option<Box<ListNode>>

Solutions

Approach 1: Elementary Math

• Time complexity: O(max(m, n))

• Space complexity: O(max(m, n))

• Runtime: 4 ms

• Memory: 2.3 MB

// 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 {
    pub fn add_two_numbers(l1: Option<Box<ListNode>>, l2: Option<Box<ListNode>>) -> Option<Box<ListNode>> {
        let mut dummy1 = l1;
        let mut dummy2 = l2;
        let mut root = Some(Box::new(ListNode::new(0)));
        let mut curr = &mut root;
        let mut carry = 0;

        while dummy1.is_some() || dummy2.is_some() {
            match curr {
                Some(inner_node) => {
                    let first = dummy1.take().unwrap_or(Box::new(ListNode::new(0)));
                    let second = dummy2.take().unwrap_or(Box::new(ListNode::new(0)));
                    let mut sum = first.val + second.val + carry;
                    carry = sum / 10;
                    sum = sum % 10;
                    inner_node.next.get_or_insert(Box::new(ListNode::new(sum)));
                    curr = &mut inner_node.next;
                    dummy1 = first.next;
                    dummy2 = second.next;
                },
                None => break,
            }
        }

        if carry == 1 {
            if let Some(node) = curr {
                node.next.get_or_insert(Box::new(ListNode::new(1)));
            }
        }

        root.unwrap().next
    }
}