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25. Reverse Nodes in k-Group

Difficulty Topics

Description

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?

Solution

reverse-nodes-in-k-group.py
# 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]:
        res = dummy = ListNode(-1)
        dummy.next = l = r = head

        while True:
            count = 0
            while r and count < k:
                r = r.next
                count += 1

            if count == k:
                prev, curr = r, l
                for _ in range(k):
                    nxt = curr.next
                    curr.next = prev
                    prev = curr
                    curr = nxt

                dummy.next, dummy, l = prev, l, r
            else:
                return res.next