# [21. 合并两个有序链表](https://leetcode-cn.com/problems/merge-two-sorted-lists)

[English Version](/solution/0000-0099/0021.Merge%20Two%20Sorted%20Lists/README_EN.md)

## 题目描述

<!-- 这里写题目描述 -->

<p>将两个升序链表合并为一个新的 <strong>升序</strong> 链表并返回。新链表是通过拼接给定的两个链表的所有节点组成的。 </p>

<p> </p>

<p><strong>示例 1：</strong></p>
<img alt="" src="https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/0000-0099/0021.Merge%20Two%20Sorted%20Lists/images/merge_ex1.jpg" style="width: 662px; height: 302px;" />
<pre>
<strong>输入：</strong>l1 = [1,2,4], l2 = [1,3,4]
<strong>输出：</strong>[1,1,2,3,4,4]
</pre>

<p><strong>示例 2：</strong></p>

<pre>
<strong>输入：</strong>l1 = [], l2 = []
<strong>输出：</strong>[]
</pre>

<p><strong>示例 3：</strong></p>

<pre>
<strong>输入：</strong>l1 = [], l2 = [0]
<strong>输出：</strong>[0]
</pre>

<p> </p>

<p><strong>提示：</strong></p>

<ul>
	<li>两个链表的节点数目范围是 <code>[0, 50]</code></li>
	<li><code>-100 <= Node.val <= 100</code></li>
	<li><code>l1</code> 和 <code>l2</code> 均按 <strong>非递减顺序</strong> 排列</li>
</ul>

## 解法

<!-- 这里可写通用的实现逻辑 -->

迭代遍历两链表，比较节点值 val 的大小，进行节点串联，得到最终链表。

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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 mergeTwoLists(self, l1: ListNode, l2: ListNode) -> ListNode:
        dummy = ListNode()
        cur = dummy
        while l1 and l2:
            if l1.val <= l2.val:
                cur.next = l1
                l1 = l1.next
            else:
                cur.next = l2
                l2 = l2.next
            cur = cur.next
        cur.next = l1 or l2
        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 mergeTwoLists(ListNode l1, ListNode l2) {
        ListNode dummy = new ListNode(0);
        ListNode cur = dummy;
        while (l1 != null && l2 != null) {
            if (l1.val <= l2.val) {
                cur.next = l1;
                l1 = l1.next;
            } else {
                cur.next = l2;
                l2 = l2.next;
            }
            cur = cur.next;
        }
        cur.next = l1 == null ? l2 : l1;
        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* mergeTwoLists(ListNode* l1, ListNode* l2) {
        ListNode* dummy = new ListNode();
        ListNode* cur = dummy;
        while (l1 && l2) {
            if (l1->val <= l2->val) {
                cur->next = l1;
                l1 = l1->next;
            } else {
                cur->next = l2;
                l2 = l2->next;
            }
            cur = cur->next;
        }
        cur->next = l1 ? l1 : l2;
        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 mergeTwoLists = function (l1, l2) {
    const dummy = new ListNode();
    let cur = dummy;
    while (l1 && l2) {
        if (l1.val <= l2.val) {
            cur.next = l1;
            l1 = l1.next;
        } else {
            cur.next = l2;
            l2 = l2.next;
        }
        cur = cur.next;
    }
    cur.next = l1 || l2;
    return dummy.next;
};
```

### **Go**

```go
/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func mergeTwoLists(l1 *ListNode, l2 *ListNode) *ListNode {
    dummy := &ListNode{}
    cur := dummy
    for l1 != nil && l2 != nil {
        if l1.Val <= l2.Val {
            cur.Next = l1
            l1 = l1.Next
        } else {
            cur.Next = l2
            l2 = l2.Next
        }
        cur = cur.Next
    }
    if l1 != nil {
        cur.Next = l1
    } else if l2 != nil {
        cur.Next = l2
    }
    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 merge_two_lists(l1, l2)
    dummy = ListNode.new()
    cur = dummy
    while l1 && l2
        if l1.val <= l2.val
            cur.next = l1
            l1 = l1.next
        else
            cur.next = l2
            l2 = l2.next
        end
        cur = cur.next
    end
    cur.next = l1 || l2
    dummy.next
end
```

### **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 MergeTwoLists(ListNode l1, ListNode l2) {
        ListNode dummy = new ListNode();
        ListNode cur = dummy;
        while (l1 != null && l2 != null) {
            if (l1.val <= l2.val) {
                cur.next = l1;
                l1 = l1.next;
            } else {
                cur.next = l2;
                l2 = l2.next;
            }
            cur = cur.next;
        }
        cur.next = l1 == null ? l2 : l1;
        return dummy.next;
    }
}
```

### **...**

```

```

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