LeetCode - Reverse Nodes in k-Group

Given the head of a linked list, reverse the nodes of the list k at a time, and return the 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.

You may not alter the values in the list's nodes, only nodes themselves may be changed.

 

Example 1:

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

Example 2:

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

 

Constraints:

• The number of nodes in the list is n.
• 1 <= k <= n <= 5000
• 0 <= Node.val <= 1000

 

My Solution:

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution(object):
    def reverseKGroup(self, head, k):
        """
        :type head: Optional[ListNode]
        :type k: int
        :rtype: Optional[ListNode]
        """

        if k == 1:
            return head
        
        def reverseBetween(head, left, right):
            if left == right or head.next is None:
                return head

            dummy = ListNode()
            dummy.next = head
            prev = dummy

            # Move prev to one node before the left-th node
            for _ in range(left - 1):
                prev = prev.next
            
            # Reverse the sublist from left to right
            curr = prev.next
            next_node = None
            for _ in range(right - left):
                next_node = curr.next
                curr.next = next_node.next
                next_node.next = prev.next
                prev.next = next_node
            
            return dummy.next
        
        node = head
        i = 1
        while node:
            next_node = node.next if node.next else None

            if i % k == 0:
                head = reverseBetween(head, i - k + 1, i)

            node = next_node
            i += 1
        
        return head