# 两两交换链表中的节点
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。
你不能只是单纯的改变节点内部的值，而是需要实际的进行节点交换。
示例 1：
输入：head = [1,2,3,4]
输出：[2,1,4,3]
示例 2：
输入：head = []
输出：[]
示例 3：
输入：head = [1]
输出：[1]
提示：
- 链表中节点的数目在范围
[0, 100] 内 0 <= Node.val <= 100
进阶：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。）
以下错误的选项是？
## aop
### before
```c
#include
using namespace std;
struct ListNode
{
int val;
struct ListNode *next;
ListNode() : val(0), next(nullptr){};
ListNode(int x) : val(x), next(nullptr){};
ListNode(int x, ListNode *next) : val(x), next(next){};
};
```
### after
```c
```
## 答案
```c
class Solution
{
public:
ListNode *swapPairs(ListNode *head)
{
ListNode *new_head = new ListNode(0);
new_head->next = head;
ListNode *pre = new_head;
while (head && head->next)
{
ListNode *first = head;
ListNode *second = head->next;
pre->next = second;
first->next = first->next;
second->next = first;
pre = first;
head = first->next;
}
return new_head->next;
}
};
```
## 选项
### A
```c
class Solution
{
public:
ListNode *swapPairs(ListNode *head)
{
if (head == NULL || head->next == NULL)
return head;
ListNode *dummy = new ListNode(0);
dummy->next = head;
ListNode *p = dummy;
while (p->next != NULL && p->next->next != NULL)
{
ListNode *node1 = p->next;
ListNode *node2 = node1->next;
ListNode *next = node2->next;
node2->next = node1;
node1->next = next;
p->next = node2;
p = node1;
}
ListNode *retnode = dummy->next;
delete dummy;
return retnode;
}
};
```
### B
```c
class Solution {
public:
ListNode* swapPairs(ListNode* head) {
if (head == nullptr || head->next == nullptr) {
return head;
}
ListNode *next = head->next;
head->next = swapPairs(next->next);
next->next = head;
return next;
}
};
```
### C
```c
class Solution
{
public:
ListNode *swapPairs(ListNode *head)
{
ListNode *p = new ListNode(-1);
p->next = head;
ListNode *h = p;
while (p->next && p->next->next)
{
ListNode *c = p->next;
ListNode *n = p->next->next;
p->next = c->next;
c->next = n->next;
n->next = c;
p = c;
}
return h->next;
}
};
```