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

  • Last Updated : 17 May, 2022

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. 


# Python program to merge a linked list 
# into another at alternate positions
class Node(object):
    def __init__(self, data:int): = data = None
class LinkedList(object):
    def __init__(self):
        self.head = None
    def push(self, new_data:int):
        new_node = Node(new_data) = 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:
            temp =
    # 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 =
            q_next =
            # make q_curr as next of p_curr
            # change next pointer of q_curr
   = p_next  
            # change next pointer of p_curr
   = 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.
# 2.
for i in range(8, 3, -1):
print("First Linked List:")
print("Second Linked List:")
# Merging the LLs
llist1.merge(p=llist1, q=llist2)
print("Modified first linked list:")
print("Modified second linked list:")
# This code is contributed by Deepanshu Mehta


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!

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