语言实例比较 2. 两数之和 C++ Rust Java Python

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 <= Node.val <= 9
题目数据保证列表表示的数字不含前导零

C++: 32ms, 70.12MB

/*** Definition for singly-linked list.* struct ListNode {*     int val;*     ListNode *next;*     ListNode() : val(0), next(nullptr) {}*     ListNode(int x) : val(x), next(nullptr) {}*     ListNode(int x, ListNode *next) : val(x), next(next) {}* };*/
class Solution {
public:ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {ListNode *dummyHead = new ListNode();ListNode *cur = dummyHead;int sum = 0;while(l1 || l2 || sum){if(l1) {sum += l1->val;l1 = l1->next;}if(l2) {sum += l2->val;l2 = l2->next;}cur->next = new ListNode(sum % 10, nullptr);cur = cur->next;sum = sum / 10;}return dummyHead->next;}
};

Rust: 方法1 递归 4ms, 2.2MB mem

// 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>> {carried(l1, l2, 0)}
}pub fn carried(l1: Option<Box<ListNode>>, l2: Option<Box<ListNode>>, mut carry: i32) -> Option<Box<ListNode>> {if l1.is_none() && l2.is_none() && carry == 0 {None} else {Some(Box::new(ListNode {next: carried(l1.and_then(|x| {carry += x.val; x.next}), l2.and_then(|x| {carry += x.val; x.next}), carry / 10),val: carry % 10}))}
}

Rust: 方法2 迭代 4ms, 2.12MB

// 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 l1 = l1;let mut l2 = l2;let mut dummy = ListNode::new(0); // 哨兵节点let mut cur = &mut dummy;let mut sum = 0; // 剩余和，进位while l1.is_some() || l2.is_some() || sum != 0 {l1 = l1.and_then(|node| {sum += node.val; node.next});l2 = l2.and_then(|node| {sum += node.val; node.next});cur.next = Some(Box::new(ListNode::new(sum % 10))); // 每个节点保存一个数位cur = cur.next.as_mut().unwrap(); // 下一个节点sum = sum / 10; // 新的进位}dummy.next // 哨兵节点的下一个节点就是头节点}
}

Java: 2ms, 43.14MB mem

/*** Definition for singly-linked list.* public class ListNode {*     int val;*     ListNode next;*     ListNode() {}*     ListNode(int val) { this.val = val; }*     ListNode(int val, ListNode next) { this.val = val; this.next = next; }* }*/
class Solution {public ListNode addTwoNumbers(ListNode l1, ListNode l2) {ListNode dummy = new ListNode();ListNode cur = dummy;int sum = 0;while(l1 != null || l2 != null || sum != 0) {if (l1 != null) { sum += l1.val; l1 = l1.next; }if (l2 != null) { sum += l2.val; l2 = l2.next; }cur.next = new ListNode(sum % 10);cur = cur.next;sum /= 10;}return dummy.next;}
}

Python: 56ms, 17.5MB mem

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:def addTwoNumbers(self, l1: Optional[ListNode], l2: Optional[ListNode]) -> Optional[ListNode]:cur = dummy = ListNode()sum = 0while l1 or l2 or sum:if l1: l1, sum = l1.next, sum + l1.valif l2: l2, sum = l2.next, sum + l2.valcur.next = ListNode(sum % 10)cur = cur.nextsum //= 10return dummy.next