Convert Linked List by Alternating Nodes from Front and Back in Python

Suppose we have a singly linked list, we have to rearrange it such that we take: the last node, and then the first node, and then the second last node, and then the second node, and so on.

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

To solve this, we will follow these steps:

• c := node

• l := a new list

• while c is not-null, do

  • insert value of c at the end of l

  • c := next of c

  • c := node

  • while c is not null and l is non-empty, do

    • value of c := value of last element from l and delete it

    • c := next of c

    • if c is null, then

      • come out from the loop

    • value of c := value of last element from l and delete it

    • c := next of c

• return node

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):
      c = node
      l = []
      while c:
         l.append(c.val)
         c = c.next

      c = node
      while c and l:
         c.val = l.pop()
         c = c.next
         if c == None:
            break
         c.val = l.pop(0)
         c = c.next
      return node

ob = Solution()
head = make_list([1,2,3,4,5,6,7,8,9])
print_list(ob.solve(head))

Input

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

Output

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