0025: Reverse Nodes in k-Group

Problem Statement

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

Follow-up: Can you solve the problem in O(1) extra memory space?

Code Solution

class Solution:
    def reverseKGroup(self, head, k):
	    
        def revk(node, k):
            prev = None
            curr = node
            while curr and k > 0:
                temp = curr.next
                curr.next = prev
                prev = curr
                curr = temp
                k = k - 1
            return prev, curr, k == 0
		
        dummy = ListNode(0)
        
        temp = dummy
        while head:
            done, head, completed = revk(head, k)
            temp.next = done if completed else revk(done, k)[0]
            while temp and temp.next:
                temp = temp.next
        return dummy.next