# 两两交换链表中的节点

<p>给定一个链表，两两交换其中相邻的节点，并返回交换后的链表。</p><p><strong>你不能只是单纯的改变节点内部的值</strong>，而是需要实际的进行节点交换。</p><p> </p><p><strong>示例 1：</strong></p><img alt="" src="https://img-service.csdnimg.cn/img_convert/88da15e3bf83dfb0c4784dc6700e84bc.jpeg" style="width: 422px; height: 222px;" /><pre><strong>输入：</strong>head = [1,2,3,4]<strong><br />输出：</strong>[2,1,4,3]</pre><p><strong>示例 2：</strong></p><pre><strong>输入：</strong>head = []<strong><br />输出：</strong>[]</pre><p><strong>示例 3：</strong></p><pre><strong>输入：</strong>head = [1]<strong><br />输出：</strong>[1]</pre><p> </p><p><strong>提示：</strong></p><ul>	<li>链表中节点的数目在范围 <code>[0, 100]</code> 内</li>	<li><code>0 <= Node.val <= 100</code></li></ul><p> </p><p><strong>进阶：</strong>你能在不修改链表节点值的情况下解决这个问题吗?（也就是说，仅修改节点本身。）</p>

以下程序实现了这一功能，请你填补空白处内容：

```cpp
#include <bits/stdc++.h>
using namespace std;
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 *swapPairs(ListNode *head)
	{
		struct ListNode dummy, *prev = &dummy, *p = head;
		dummy.next = head;
		while (p != nullptr && p->next != nullptr)
		{
			struct ListNode *q = p->next;
			_____________________;
		}
		return dummy.next;
	}
};
```

## template

```cpp
#include <bits/stdc++.h>
using namespace std;
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 *swapPairs(ListNode *head)
	{
		struct ListNode dummy, *prev = &dummy, *p = head;
		dummy.next = head;
		while (p != nullptr && p->next != nullptr)
		{
			struct ListNode *q = p->next;
			p->next = q->next;
			q->next = prev->next;
			prev->next = q;
			prev = p;
			p = p->next;
		}
		return dummy.next;
	}
};
```

## 答案

```cpp
p->next = q->next;
q->next = prev->next;
prev->next = q;
prev = p;
p = p->next;
```

## 选项

### A

```cpp
q->next = prev->next;
p->next = q->next;
prev->next = q;
prev = p;
p = p->next;
```

### B

```cpp
prev->next = q;
p->next = q->next;
q->next = prev->next;
prev = p;
p = p->next;
```

### C

```cpp
p->next = q->next;
prev->next = q;
prev = p;
q->next = prev->next;
p = p->next;
```