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# 25.reverse-nodes-in-k-groups-cn

## 题目地址

<https://leetcode.com/problems/reverse-nodes-in-k-group/>

## 题目描述

```
Given a linked list, reverse the nodes of a linked list k at a time and return its modified list.

k is a positive integer and is less than or equal to the length of the linked list. If the number of nodes is not a multiple of k then left-out nodes in the end should remain as it is.

Example:

Given this linked list: 1->2->3->4->5

For k = 2, you should return: 2->1->4->3->5

For k = 3, you should return: 3->2->1->4->5

Note:

Only constant extra memory is allowed.
You may not alter the values in the list's nodes, only nodes itself may be changed.
```

## 思路

题意是以 `k` 个nodes为一组进行翻转，返回翻转后的`linked list`.

从左往右扫描一遍`linked list`，扫描过程中，以k为单位把数组分成若干段，对每一段进行翻转。给定首尾nodes，如何对链表进行翻转。

链表的翻转过程，初始化一个为`null`的 `previous node（prev）`，然后遍历链表的同时，当前`node （curr）`的下一个（next）指向前一个`node（prev）`， 在改变当前node的指向之前，用一个临时变量记录当前node的下一个`node（curr.next)`. 即

```
ListNode temp = curr.next;
curr.next = prev;
prev = curr;
curr = temp;
```

举例如图：翻转整个链表 `1->2->3->4->null` -> `4->3->2->1->null`

![reverse linked list](/files/-Lni143mEUS5xjd0V4nS)

这里是对每一组（`k个nodes`）进行翻转，

1. 先分组，用一个`count`变量记录当前节点的个数
2. 用一个`start` 变量记录当前分组的起始节点位置的前一个节点
3. 用一个`end`变量记录要翻转的最后一个节点位置
4. 翻转一组（`k个nodes`）即`(start, end) - start and end exclusively`。
5. 翻转后，`start`指向翻转后链表, 区间`（start，end）`中的最后一个节点, 返回`start` 节点。
6. 如果不需要翻转，`end` 就往后移动一个（`end=end.next`)，每一次移动，都要`count+1`.

如图所示 步骤4和5： 翻转区间链表区间`（start， end）`

![reverse linked list range in (start, end)](/files/-Lnkyi2sjCLXYHMouDo5)

举例如图，`head=[1,2,3,4,5,6,7,8], k = 3`

![reverse k nodes in linked list](/files/-Lni143oh3jIlcvc0ANx)

> **NOTE**: 一般情况下对链表的操作，都有可能会引入一个新的`dummy node`，因为`head`有可能会改变。这里`head 从1->3`, `dummy (List(0))`保持不变。

### 复杂度分析

* *时间复杂度:* `O(n) - n is number of Linked List`
* *空间复杂度:* `O(1)`

## 关键点分析

1. 创建一个dummy node
2. 对链表以k为单位进行分组，记录每一组的起始和最后节点位置
3. 对每一组进行翻转，更换起始和最后的位置
4. 返回`dummy.next`.

## 代码 (`Java/Python3`)

*Java Code*

```java
class ReverseKGroupsLinkedList {
  public ListNode reverseKGroup(ListNode head, int k) {
      if (head == null || k == 1) {
        return head;
      }
      ListNode dummy = new ListNode(0);
      dummy.next = head;

      ListNode start = dummy;
      ListNode end = head;
      int count = 0;
      while (end != null) {
        count++;
        // group
        if (count % k == 0) {
          // reverse linked list (start, end]
          start = reverse(start, end.next);
          end = start.next;
        } else {
          end = end.next;
        }
      }
      return dummy.next;
    }

    /** 
     * reverse linked list from range (start, end), return last node.
     * for example: 
     * 0->1->2->3->4->5->6->7->8
     * |           |
     * start       end
     * 
     * After call start = reverse(start, end)
     * 
     * 0->3->2->1->4->5->6->7->8
     *          |  |
     *       start end
     *       first
     * @return the reversed list's 'start' node, which is the precedence of node end
     */
    private ListNode reverse(ListNode start, ListNode end) {
      ListNode curr = start.next;
      ListNode prev = start;
      ListNode first = curr;
      while (curr != end){
        ListNode temp = curr.next;
        curr.next = prev;
        prev = curr;
        curr = temp;
      }
      start.next = prev;
      first.next = curr;
      return first;
    }
}
```

*Python3 Cose*

```python
class Solution:
    def reverseKGroup(self, head: ListNode, k: int) -> ListNode:
        if head is None or k < 2:
            return head
        dummy = ListNode(0)
        dummy.next = head
        start = dummy
        end = head
        count = 0
        while end:
            count += 1
            if count % k == 0:
                start = self.reverse(start, end.next)
                end = start.next
            else:
                end = end.next
        return dummy.next

    def reverse(self, start, end):
        prev, curr = start, start.next
        first = curr
        while curr != end:
            temp = curr.next
            curr.next = prev
            prev = curr
            curr = temp
        start.next = prev
        first.next = curr
        return first
```

## 参考（References)

* [Leetcode Discussion (yellowstone)](https://leetcode.com/problems/reverse-nodes-in-k-group/discuss/11440/Non-recursive-Java-solution-and-idea)

## 扩展

* 要求从后往前以`k`个为一组进行翻转。**(字节跳动（ByteDance）面试题)**

  例子，`1->2->3->4->5->6->7->8, k = 3`,

  从后往前以`k=3`为一组，

  * `6->7->8` 为一组翻转为`8->7->6`，&#x20;
  * `3->4->5`为一组翻转为`5->4->3`.&#x20;
  * `1->2`只有2个nodes少于`k=3`个，不翻转。

    最后返回： `1->2->5->4->3->8->7->6`

这里的思路跟从前往后以`k`个为一组进行翻转类似，可以进行预处理：

1. 翻转链表
2. 对翻转后的链表进行从前往后以k为一组翻转。
3. 翻转步骤2中得到的链表。

例子：`1->2->3->4->5->6->7->8, k = 3`

1. 翻转链表得到：`8->7->6->5->4->3->2->1`
2. 以k为一组翻转： `6->7->8->3->4->5->2->1`
3. 翻转步骤#2链表： `1->2->5->4->3->8->7->6`

## 类似题目

* [Swap Nodes in Pairs](https://leetcode.com/problems/swap-nodes-in-pairs/)