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p0002_add_two_numbers.rs
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/**
* [2] Add Two Numbers
*
* You are given two non-empty linked lists representing two non-negative integers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list.
* You may assume the two numbers do not contain any leading zero, except the number 0 itself.
* Example:
*
* Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
* Output: 7 -> 0 -> 8
* Explanation: 342 + 465 = 807.
*
*/
pub struct Solution {}
use super::utils::linked_list::{ListNode, to_list};
use std::borrow::BorrowMut;
// submission codes
impl Solution {
pub fn add_two_numbers(l1: Option<Box<ListNode>>, l2: Option<Box<ListNode>>) -> Option<Box<ListNode>> {
let (mut l1, mut l2) = (l1, l2);
let mut sum = 0;
let mut front = ListNode::new(0);
let mut current = &mut front;
loop {
let mut not_end_count = 0;
if let Some(node) = l1 {
sum += node.val;
not_end_count += 1;
l1 = node.next;
}
if let Some(node) = l2 {
sum += node.val;
not_end_count += 1;
l2 = node.next;
}
if not_end_count > 0 || sum > 0 {
current.next = Some(Box::new(ListNode::new(sum % 10)));
// current = current.next.as_deref_mut().unwrap(); // leetcode不支持
current = current.next.as_mut().unwrap().as_mut();
sum /= 10;
} else {
break front.next
}
}
}
}
// submission codes end
#[cfg(test)]
mod tests {
use super::*;
#[test]
fn test_2() {
assert_eq!(Solution::add_two_numbers(to_list(vec![2, 4, 3]), to_list(vec![5, 6, 4])), to_list(vec![7, 0, 8]));
assert_eq!(Solution::add_two_numbers(to_list(vec![1]), to_list(vec![1])), to_list(vec![2]));
assert_eq!(Solution::add_two_numbers(to_list(vec![9]), to_list(vec![9])), to_list(vec![8, 1]));
assert_eq!(Solution::add_two_numbers(to_list(vec![9]), to_list(vec![1, 2])), to_list(vec![0, 3]));
assert_eq!(Solution::add_two_numbers(to_list(vec![0]), to_list(vec![0])), to_list(vec![0]))
}
}