# [2. 两数相加](https://leetcode.cn/problems/add-two-numbers)
[English Version](/solution/0000-0099/0002.Add%20Two%20Numbers/README_EN.md)
## 题目描述
<!-- 这里写题目描述 -->
<p>给你两个 <strong>非空</strong> 的链表，表示两个非负的整数。它们每位数字都是按照 <strong>逆序</strong> 的方式存储的，并且每个节点只能存储 <strong>一位</strong> 数字。</p>
<p>请你将两个数相加，并以相同形式返回一个表示和的链表。</p>
<p>你可以假设除了数字 0 之外，这两个数都不会以 0 开头。</p>
<p> </p>
<p><strong>示例 1：</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0000-0099/0002.Add%20Two%20Numbers/images/addtwonumber1.jpg" style="width: 483px; height: 342px;" />
<pre>
<strong>输入：</strong>l1 = [2,4,3], l2 = [5,6,4]
<strong>输出：</strong>[7,0,8]
<strong>解释：</strong>342 + 465 = 807.
</pre>
<p><strong>示例 2：</strong></p>
<pre>
<strong>输入：</strong>l1 = [0], l2 = [0]
<strong>输出：</strong>[0]
</pre>
<p><strong>示例 3：</strong></p>
<pre>
<strong>输入：</strong>l1 = [9,9,9,9,9,9,9], l2 = [9,9,9,9]
<strong>输出：</strong>[8,9,9,9,0,0,0,1]
</pre>
<p> </p>
<p><strong>提示：</strong></p>
<ul>
<li>每个链表中的节点数在范围 <code>[1, 100]</code> 内</li>
<li><code>0 <= Node.val <= 9</code></li>
<li>题目数据保证列表表示的数字不含前导零</li>
</ul>
## 解法
<!-- 这里可写通用的实现逻辑 -->
**方法一：模拟**
我们同时遍历两个链表 $l_1$ 和 $l_2$，并使用变量 $carry$ 表示当前是否有进位。
每次遍历时，我们取出对应链表的当前位，计算它们与进位 $carry$ 的和，然后更新进位的值，最后将当前位的值加入答案链表。如果两个链表都遍历完了，并且进位为 $0$ 时，遍历结束。
最后我们返回答案链表的头节点即可。
时间复杂度 $O(max(m, n))$，其中 $m$ 和 $n$ 分别为两个链表的长度。我们需要遍历两个链表的全部位置，而处理每个位置只需要 $O(1)$ 的时间。忽略答案的空间消耗，空间复杂度 $O(1)$。
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### **Python3**
<!-- 这里可写当前语言的特殊实现逻辑 -->
```python
# 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]:
dummy = ListNode()
carry, curr = 0, dummy
while l1 or l2 or carry:
s = (l1.val if l1 else 0) + (l2.val if l2 else 0) + carry
carry, val = divmod(s, 10)
curr.next = ListNode(val)
curr = curr.next
l1 = l1.next if l1 else None
l2 = l2.next if l2 else None
return dummy.next
```
### **Java**
<!-- 这里可写当前语言的特殊实现逻辑 -->
```java
/**
* 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(0);
int carry = 0;
ListNode cur = dummy;
while (l1 != null || l2 != null || carry != 0) {
int s = (l1 == null ? 0 : l1.val) + (l2 == null ? 0 : l2.val) + carry;
carry = s / 10;
cur.next = new ListNode(s % 10);
cur = cur.next;
l1 = l1 == null ? null : l1.next;
l2 = l2 == null ? null : l2.next;
}
return dummy.next;
}
}
```
### **C++**
```cpp
/**
* 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* dummy = new ListNode();
int carry = 0;
ListNode* cur = dummy;
while (l1 || l2 || carry) {
int s = (l1 ? l1->val : 0) + (l2 ? l2->val : 0) + carry;
carry = s / 10;
cur->next = new ListNode(s % 10);
cur = cur->next;
l1 = l1 ? l1->next : nullptr;
l2 = l2 ? l2->next : nullptr;
}
return dummy->next;
}
};
```
### **JavaScript**
```js
/**
* Definition for singly-linked list.
* function ListNode(val, next) {
* this.val = (val===undefined ? 0 : val)
* this.next = (next===undefined ? null : next)
* }
*/
/**
* @param {ListNode} l1
* @param {ListNode} l2
* @return {ListNode}
*/
var addTwoNumbers = function (l1, l2) {
const dummy = new ListNode();
let carry = 0;
let cur = dummy;
while (l1 || l2 || carry) {
const s = (l1?.val || 0) + (l2?.val || 0) + carry;
carry = Math.floor(s / 10);
cur.next = new ListNode(s % 10);
cur = cur.next;
l1 = l1?.next;
l2 = l2?.next;
}
return dummy.next;
};
```
### **C#**
```cs
/**
* Definition for singly-linked list.
* public class ListNode {
* public int val;
* public ListNode next;
* public ListNode(int val=0, ListNode next=null) {
* this.val = val;
* this.next = next;
* }
* }
*/
public class Solution {
public ListNode AddTwoNumbers(ListNode l1, ListNode l2) {
ListNode dummy = new ListNode();
int carry = 0;
ListNode cur = dummy;
while (l1 != null || l2 != null || carry != 0) {
int s = (l1 == null ? 0 : l1.val) + (l2 == null ? 0 : l2.val) + carry;
carry = s / 10;
cur.next = new ListNode(s % 10);
cur = cur.next;
l1 = l1 == null ? null : l1.next;
l2 = l2 == null ? null : l2.next;
}
return dummy.next;
}
}
```
### **Go**
```go
/**
* Definition for singly-linked list.
* type ListNode struct {
* Val int
* Next *ListNode
* }
*/
func addTwoNumbers(l1 *ListNode, l2 *ListNode) *ListNode {
dummy := &ListNode{}
carry := 0
cur := dummy
for l1 != nil || l2 != nil || carry != 0 {
s := carry
if l1 != nil {
s += l1.Val
}
if l2 != nil {
s += l2.Val
}
carry = s / 10
cur.Next = &ListNode{s % 10, nil}
cur = cur.Next
if l1 != nil {
l1 = l1.Next
}
if l2 != nil {
l2 = l2.Next
}
}
return dummy.Next
}
```
### **Ruby**
```rb
# Definition for singly-linked list.
# class ListNode
# attr_accessor :val, :next
# def initialize(val = 0, _next = nil)
# @val = val
# @next = _next
# end
# end
# @param {ListNode} l1
# @param {ListNode} l2
# @return {ListNode}
def add_two_numbers(l1, l2)
dummy = ListNode.new()
carry = 0
cur = dummy
while !l1.nil? || !l2.nil? || carry > 0
s = (l1.nil? ? 0 : l1.val) + (l2.nil? ? 0 : l2.val) + carry
carry = s / 10
cur.next = ListNode.new(s % 10)
cur = cur.next
l1 = l1.nil? ? l1 : l1.next
l2 = l2.nil? ? l2 : l2.next
end
dummy.next
end
```
### **Swift**
```swift
/**
* Definition for singly-linked list.
* public class ListNode {
* public var val: Int
* public var next: ListNode?
* public init() { self.val = 0; self.next = nil; }
* public init(_ val: Int) { self.val = val; self.next = nil; }
* public init(_ val: Int, _ next: ListNode?) { self.val = val; self.next = next; }
* }
*/
class Solution {
func addTwoNumbers(_ l1: ListNode?, _ l2: ListNode?) -> ListNode? {
var dummy = ListNode.init()
var carry = 0
var l1 = l1
var l2 = l2
var cur = dummy
while l1 != nil || l2 != nil || carry != 0 {
let s = (l1?.val ?? 0) + (l2?.val ?? 0) + carry
carry = s / 10
cur.next = ListNode.init(s % 10)
cur = cur.next!
l1 = l1?.next
l2 = l2?.next
}
return dummy.next
}
}
```
### **Nim**
```nim
#[
# Driver code in the solution file
# Definition for singly-linked list.
type
Node[int] = ref object
value: int
next: Node[int]
SinglyLinkedList[T] = object
head, tail: Node[T]
]#
# More efficient code churning ...
proc addTwoNumbers(l1: var SinglyLinkedList, l2: var SinglyLinkedList): SinglyLinkedList[int] =
var
aggregate: SinglyLinkedList
psum: seq[char]
temp_la, temp_lb: seq[int]
while not l1.head.isNil:
temp_la.add(l1.head.value)
l1.head = l1.head.next
while not l2.head.isNil:
temp_lb.add(l2.head.value)
l2.head = l2.head.next
psum = reversed($(reversed(temp_la).join("").parseInt() + reversed(temp_lb).join("").parseInt()))
for i in psum: aggregate.append(($i).parseInt())
result = aggregate
```
### **TypeScript**
```ts
/**
* Definition for singly-linked list.
* class ListNode {
* val: number
* next: ListNode | null
* constructor(val?: number, next?: ListNode | null) {
* this.val = (val===undefined ? 0 : val)
* this.next = (next===undefined ? null : next)
* }
* }
*/
function addTwoNumbers(l1: ListNode | null, l2: ListNode | null): ListNode | null {
const dummy = new ListNode();
let cur = dummy;
let 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 = Math.floor(sum / 10);
}
return dummy.next;
}
```
### **Rust**
```rust
// 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(
mut l1: Option<Box<ListNode>>,
mut l2: Option<Box<ListNode>>,
) -> Option<Box<ListNode>> {
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;
}
dummy.unwrap().next.take()
}
}
```
### **...**
```
```
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