# K 个一组翻转链表
给你一个链表，每 k 个节点一组进行翻转，请你返回翻转后的链表。
k 是一个正整数，它的值小于或等于链表的长度。
如果节点总数不是 k 的整数倍，那么请将最后剩余的节点保持原有顺序。
进阶：
- 你可以设计一个只使用常数额外空间的算法来解决此问题吗？
- 你不能只是单纯的改变节点内部的值，而是需要实际进行节点交换。
示例 1：
输入：head = [1,2,3,4,5], k = 2
输出：[2,1,4,3,5]
示例 2：
输入：head = [1,2,3,4,5], k = 3
输出：[3,2,1,4,5]
示例 3：
输入：head = [1,2,3,4,5], k = 1
输出：[1,2,3,4,5]
示例 4：
输入：head = [1], k = 1
输出：[1]
提示：
- 列表中节点的数量在范围
sz 内 1 <= sz <= 50000 <= Node.val <= 10001 <= k <= sz
以下程序实现了这一功能，请你填补空白处的内容：
```cpp
#include
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 *reverseGroup(ListNode *head, int k)
{
int len = 0;
struct ListNode dummy, *prev = &dummy;
dummy.next = head;
for (; head != nullptr; head = head->next)
{
____________________;
}
return dummy.next;
}
};
```
## template
```cpp
#include
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 *reverseGroup(ListNode *head, int k)
{
int len = 0;
struct ListNode dummy, *prev = &dummy;
dummy.next = head;
for (; head != nullptr; head = head->next)
{
if (++len % k == 0)
{
struct ListNode *p = prev->next;
while (prev->next != head)
{
struct ListNode *q = p->next;
p->next = q->next;
q->next = prev->next;
prev->next = q;
}
prev = p;
head = p;
}
}
return dummy.next;
}
};
```
## 答案
```cpp
if (++len % k == 0)
{
struct ListNode *p = prev->next;
while (prev->next != head)
{
struct ListNode *q = p->next;
p->next = q->next;
q->next = prev->next;
prev->next = q;
}
prev = p;
head = p;
}
```
## 选项
### A
```cpp
if (len % k == 0)
{
struct ListNode *p = prev->next;
while (prev->next != head)
{
struct ListNode *q = p->next;
p->next = q->next;
q->next = prev->next;
prev->next = q;
}
prev = p;
head = p;
}
```
### B
```cpp
if (++len % k == 0)
{
struct ListNode *p = prev->next;
while (prev->next != head)
{
struct ListNode *q = p->next;
prev->next = q;
q->next = prev->next;
p->next = q->next;
}
prev = p;
head = p;
}
```
### C
```cpp
if (++len % k == 0)
{
struct ListNode *p = prev->next;
while (prev->next != head)
{
struct ListNode *q = p->next;
prev->next = q;
p->next = q->next;
q->next = prev->next;
}
prev = p;
head = p;
}
```