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2.两数相加.rs
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/*
* @lc app=leetcode.cn id=2 lang=rust
*
* [2] 两数相加
*
* https://leetcode-cn.com/problems/add-two-numbers/description/
*
* algorithms
* Medium (40.96%)
* Likes: 6937
* Dislikes: 0
* Total Accepted: 1M
* Total Submissions: 2.5M
* Testcase Example: '[2,4,3]\n[5,6,4]'
*
* 给你两个 非空 的链表,表示两个非负的整数。它们每位数字都是按照 逆序 的方式存储的,并且每个节点只能存储 一位 数字。
*
* 请你将两个数相加,并以相同形式返回一个表示和的链表。
*
* 你可以假设除了数字 0 之外,这两个数都不会以 0 开头。
*
*
*
* 示例 1:
*
*
* 输入:l1 = [2,4,3], l2 = [5,6,4]
* 输出:[7,0,8]
* 解释:342 + 465 = 807.
*
*
* 示例 2:
*
*
* 输入:l1 = [0], l2 = [0]
* 输出:[0]
*
*
* 示例 3:
*
*
* 输入:l1 = [9,9,9,9,9,9,9], l2 = [9,9,9,9]
* 输出:[8,9,9,9,0,0,0,1]
*
*
*
*
* 提示:
*
*
* 每个链表中的节点数在范围 [1, 100] 内
* 0
* 题目数据保证列表表示的数字不含前导零
*
*
*/
// @lc code=start
// 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 result = None;
let mut tail = &mut result;
let mut t = (l1, l2, 0, 0); // (list1, list2, sum, carry)
loop {
t = match t {
(None, None, _, 0) => break,
(None, None, _, carry) => (None, None, carry, 0),
(Some(list), None, _, carry) | (None, Some(list), _, carry) if list.val + carry >= 10 => {
(list.next, None, list.val + carry - 10, 1)
}
(Some(list), None, _, carry) | (None, Some(list), _, carry) => {
(list.next, None, list.val + carry, 0)
}
(Some(l1), Some(l2), _, carry) if l1.val + l2.val + carry >= 10 => {
(l1.next, l2.next, l1.val + l2.val + carry - 10, 1)
}
(Some(l1), Some(l2), _, carry) => {
(l1.next, l2.next, l1.val + l2.val + carry, 0)
}
};
*tail = Some(Box::new(ListNode::new(t.2)));
tail = & mut tail.as_mut().unwrap().next;
}
result
}
}
// @lc code=end