Python Program To Merge A Linked List Into Another Linked List At Alternate Positions

Given two linked lists, insert nodes of the second list into the first list at alternate positions of the first list. 
For example, if first list is 5->7->17->13->11 and second is 12->10->2->4->6, the first list should become 5->12->7->10->17->2->13->4->11->6 and second list should become empty. The nodes of the second list should only be inserted when there are positions available. For example, if the first list is 1->2->3 and the second list is 4->5->6->7->8, then the first list should become 1->4->2->5->3->6 and the second list to 7->8.
Use of extra space is not allowed (Not allowed to create additional nodes), i.e., insertion must be done in-place. The expected time complexity is O(n) where n is the number of nodes in the first list.

The idea is to run a loop while there are available positions in first loop and insert nodes of second list by changing pointers. Following are implementations of this approach. 

Python3

# Python program to merge a linked list
# into another at alternate positions
class Node(object):
    def __init__(self, data:int):
        self.data = data
        self.next = None

class LinkedList(object):
    def __init__(self):
        self.head = None

    def push(self, new_data:int):
        new_node = Node(new_data)
        new_node.next = self.head

        # 4. Move the head to point to
        # new Node
        self.head = new_node

    # Function to print linked list from
    # the Head
    def printList(self):
        temp = self.head
        while temp != None:
            print(temp.data)
            temp = temp.next

    # Main function that inserts nodes of linked
    # list q into p at alternate positions.
    # Since head of first list never changes
    # but head of second list/ may change,
    # we need single pointer for first list and
    # double pointer for second list.
    def merge(self, p, q):
        p_curr = p.head
        q_curr = q.head

        # swap their positions until one
        # finishes off
        while p_curr != None and q_curr != None:

            # Save next pointers
            p_next = p_curr.next
            q_next = q_curr.next

            # make q_curr as next of p_curr
            # change next pointer of q_curr
            q_curr.next = p_next

            # change next pointer of p_curr
            p_curr.next = q_curr

            # update current pointers for next
            # iteration
            p_curr = p_next
            q_curr = q_next
            q.head = q_curr

# Driver code
llist1 = LinkedList()
llist2 = LinkedList()

# Creating Linked lists
# 1.
llist1.push(3)
llist1.push(2)
llist1.push(1)
llist1.push(0)

# 2.
for i in range(8, 3, -1):
    llist2.push(i)

print("First Linked List:")
llist1.printList()

print("Second Linked List:")
llist2.printList()

# Merging the LLs
llist1.merge(p=llist1, q=llist2)

print("Modified first linked list:")
llist1.printList()

print("Modified second linked list:")
llist2.printList()

# This code is contributed by Deepanshu Mehta

Output:

First Linked List:
1 2 3
Second Linked List:
4 5 6 7 8
Modified First Linked List:
1 4 2 5 3 6
Modified Second Linked List:
7 8 

Time Complexity: O(N)

Auxiliary Space: O(1)

Please refer complete article on Merge a linked list into another linked list at alternate positions for more details!