README.md

25. 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

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

Solutions (Python)

1. Solution

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def reverseKGroup(self, head: Optional[ListNode], k: int) -> Optional[ListNode]:
        length = 0
        curr = head
        while curr is not None:
            length += 1
            curr = curr.next

        dummy = ListNode(next=head)
        grouptail = dummy

        for _ in range(length // k):
            grouphead = grouptail.next
            prev = grouphead
            for _ in range(k):
                prev = prev.next

            curr = grouphead
            for _ in range(k):
                temp = curr
                curr = curr.next
                temp.next = prev
                prev = temp

            grouptail.next = prev
            grouptail = grouphead

        return dummy.next