Reverse Linked List by Groups of Size K in Python

Suppose we have a singly linked list, and another value k, we have to reverse every k contiguous group of nodes.

So, if the input is like List = [1,2,3,4,5,6,7,8,9,10], k = 3, then the output will be [3, 2, 1, 6, 5, 4, 9, 8, 7, 10, ]

To solve this, we will follow these steps −

• tmp := a new node with value 0
• next of tmp := node
• prev := null, curr := null
• lp := temp, lc := curr
• cnt := k
• while curr is not null, do
  • prev := null
  • while cnt > 0 and curr is not null, do
    • following := next of curr
    • next of curr := prev
    • prev := curr, curr := following
    • cnt := cnt - 1
  • next of lp := prev, next of lc := curr
  • lp := lc, lc := curr
  • cnt := k
• return next of tmp

Let us see the following implementation to get better understanding −

Example

class ListNode:
   def __init__(self, data, next = None):
      self.val = data
      self.next = next
def make_list(elements):
   head = ListNode(elements[0])
   for element in elements[1:]:
      ptr = head
      while ptr.next:
         ptr = ptr.next
      ptr.next = ListNode(element)
   return head
def print_list(head):
   ptr = head print('[', end = "")
   while ptr:
      print(ptr.val, end = ", ")
      ptr = ptr.next
      print(']')
class Solution:
   def solve(self, node, k):
      tmp = ListNode(0)
      tmp.next = node
      prev, curr = None, node
      lp, lc = tmp, curr
      cnt = k
      while curr:
         prev = None
         while cnt > 0 and curr:
            following = curr.next
            curr.next = prev
            prev, curr = curr, following
            cnt -= 1
         lp.next, lc.next = prev, curr
         lp, lc = lc, curr
         cnt = k
      return tmp.next
ob = Solution()
head = make_list([1,2,3,4,5,6,7,8,9,10])
print_list(ob.solve(head, 3))

Input

[1,2,3,4,5,6,7,8,9,10], 3

Output

[3, 2, 1, 6, 5, 4, 9, 8, 7, 10, ]