# 合并两个有序链表

<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://img-service.csdnimg.cn/img_convert/4c4b9ef8c8206165cf198a972bcd6088.jpeg" style="width: 662px; height: 302px;" /><pre><strong>输入：</strong>l1 = [1,2,4], l2 = [1,3,4]<strong><br />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><br />输出：</strong>[]</pre><p><strong>示例 3：</strong></p><pre><strong>输入：</strong>l1 = [], l2 = [0]<strong><br />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>

## template

```java
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 h = new ListNode(0, null);
		ListNode p = h;
		while (l1 != null && l2 != null) {
			if (l1.val < l2.val) {
				p.next = l1;
				p = l1;
				l1 = l1.next;
			} else {
				p.next = l2;
				p = l2;
				l2 = l2.next;
			}
		}
		if (l1 != null) {
			p.next = l1;
		} else {
			p.next = l2;
		}
		return h.next;
	}
}
```

## 答案

```java

```

## 选项

### A

```java

```

### B

```java

```

### C

```java

```