# 25. Reverse Nodes in k-Group

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.

# Solution

Approach 1: List manipulation.

# Code (Python)

Approach 1:

# Code (C++)

Approach 1:

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
//  dummyHead->l1->l2->l3->l4->... (k = 2)
//       |      |   |   |
//     left  begin end  fast

class Solution {
public:
    ListNode* reverseKGroup(ListNode* head, int k) {
        ListNode dummyHead(0);
        ListNode *left = &dummyHead;
        left->next = head;
        ListNode *fast = head;
        while (fast)
        {
            ListNode *begin = fast;
            ListNode *end = left;
            int i = 0;
            for (; i < k && fast; ++i)
            {
                fast = fast->next;
                end = end->next;
            }
            if (i < k)
                break;
            ListNode *prev = left;
            ListNode *curr = begin;
            while (curr != fast)
            {
                ListNode *next = curr->next;
                curr->next = prev;
                prev = curr;
                curr = next;
            }
            left->next = end;
            begin->next = fast;
            left = begin;
        }
        return dummyHead.next;
    }
};