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# LeetCode 25. Reverse Nodes in k-Group

Linked List

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.

**Follow up:**

* Could you solve the problem in `O(1)` extra memory space?
* You may not alter the values in the list's nodes, only nodes itself may be changed.

**Example 1:**![](https://assets.leetcode.com/uploads/2020/10/03/reverse_ex1.jpg)

```
Input: head = [1,2,3,4,5], k = 2
Output: [2,1,4,3,5]
```

**Example 2:**![](https://assets.leetcode.com/uploads/2020/10/03/reverse_ex2.jpg)

```
Input: head = [1,2,3,4,5], k = 3
Output: [3,2,1,4,5]
```

**Example 3:**

```
Input: head = [1,2,3,4,5], k = 1
Output: [1,2,3,4,5]
```

**Example 4:**

```
Input: head = [1], k = 1
Output: [1]
```

**Constraints:**

* The number of nodes in the list is in the range `sz`.
* `1 <= sz <= 5000`
* `0 <= Node.val <= 1000`
* `1 <= k <= sz`

## Solution:

[English Version in Youtube](https://youtu.be/dc4gxhpCrPY)

[中文版解答Youtube Link](https://youtu.be/goNpyRA02jw)

[中文版解答Bilibili Link](https://www.bilibili.com/video/BV18K4y1J7Fm/)

```
/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
    
    int length(ListNode* head) {
        int len = 0;
        while (head != nullptr) {
            head = head->next;
            len++;
        }
        return len;
    }
    
    ListNode* dfs(ListNode* head, int len, int k) {
        if (len < k) {
            return head;
        }
        ListNode* tail = head;
        
        ListNode* prev = nullptr, *tmp = nullptr;
        for (int i = 0; i < k; i++) {
            tmp = head->next;
            head -> next = prev;
            prev = head;
            head = tmp;
        }
        
        tail->next = dfs(head, len - k, k);
        return prev;
    }
    
public:
    ListNode* reverseKGroup(ListNode* head, int k) {
        return dfs(head, length(head), k);
    }
};
```