solution.md

# K 个一组翻转链表

<p>给你一个链表，每 <em>k </em>个节点一组进行翻转，请你返回翻转后的链表。</p><p><em>k </em>是一个正整数，它的值小于或等于链表的长度。</p><p>如果节点总数不是 <em>k </em>的整数倍，那么请将最后剩余的节点保持原有顺序。</p><p><strong>进阶：</strong></p><ul>	<li>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？</li>	<li><strong>你不能只是单纯的改变节点内部的值</strong>，而是需要实际进行节点交换。</li></ul><p> </p><p><strong>示例 1：</strong></p><img alt="" src="https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/0000-0099/0025.Reverse%20Nodes%20in%20k-Group/images/reverse_ex1.jpg" style="width: 542px; height: 222px;" /><pre><strong>输入：</strong>head = [1,2,3,4,5], k = 2<strong><br />输出：</strong>[2,1,4,3,5]</pre><p><strong>示例 2：</strong></p><img alt="" src="https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/0000-0099/0025.Reverse%20Nodes%20in%20k-Group/images/reverse_ex2.jpg" style="width: 542px; height: 222px;" /><pre><strong>输入：</strong>head = [1,2,3,4,5], k = 3<strong><br />输出：</strong>[3,2,1,4,5]</pre><p><strong>示例 3：</strong></p><pre><strong>输入：</strong>head = [1,2,3,4,5], k = 1<strong><br />输出：</strong>[1,2,3,4,5]</pre><p><strong>示例 4：</strong></p><pre><strong>输入：</strong>head = [1], k = 1<strong><br />输出：</strong>[1]</pre><ul></ul><p><strong>提示：</strong></p><ul>	<li>列表中节点的数量在范围 <code>sz</code> 内</li>	<li><code>1 <= sz <= 5000</code></li>	<li><code>0 <= Node.val <= 1000</code></li>	<li><code>1 <= k <= sz</code></li></ul>

以下程序实现了这一功能，请你填补空白处的内容：

```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 *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 <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 *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;
}
```