Python Program For Alternating Split Of A Given Singly Linked List- Set 1

Last Updated : 23 Jan, 2023

Write a function AlternatingSplit() that takes one list and divides up its nodes to make two smaller lists 'a' and 'b'. The sublists should be made from alternating elements in the original list. So if the original list is 0->1->0->1->0->1 then one sublist should be 0->0->0 and the other should be 1->1->1.

Method (Simple):
The simplest approach iterates over the source list and pull nodes off the source and alternately put them at the front (or beginning) of 'a' and b'. The only strange part is that the nodes will be in the reverse order that occurred in the source list. Method 2 inserts the node at the end by keeping track of the last node in sublists.

Python

# Python program to alternatively split 
# a linked list into two halves 

# Node class
class Node:    
    def __init__(self, data, 
                 next = None):        
        self.data = data
        self.next = None

class LinkedList:    
    def __init__(self):        
        self.head = None
    
    # Given the source list, split its nodes 
    # into two shorter lists. If we number the 
    # elements 0, 1, 2, ... then all the even 
    # elements should go in the first list, and 
    # all the odd elements in the second. The 
    # elements in the new lists may be in any order.
    def AlternatingSplit(self, a, b):        
        first = self.head
        second = first.next
        
        while (first is not None and 
               second is not None and 
               first.next is not None):
              
              # Move a node to list 'a'
              self.MoveNode(a, first) 
              
              # Move a node to list 'b'
              self.MoveNode(b, second) 
              
              first = first.next.next
              
              if first is None:
                break
              
              second = first.next
            
    # Pull off the front node of the 
    # source and put it in dest
    def MoveNode(self, dest, node):
        
        # Make the new node
        new_node = Node(node.data)
        
        if dest.head is None:
            dest.head = new_node
        else:
            
            # Link the old dest off the 
            # new node 
            new_node.next = dest.head
            
            # Move dest to point to the 
            # new node 
            dest.head = new_node

    # Utility Functions
    # Function to insert a node at  
    # the beginning of the linked list 
    def push(self, data):
        
        # 1 & 2 allocate the Node & 
        # put the data
        
        new_node = Node(data)
        
        # Make the next of new Node as head
        new_node.next = self.head
        
        # Move the head to point to 
        # new Node
        self.head = new_node
        
    # Function to print nodes in a given 
    # linked list 
    def printList(self):
        
        temp = self.head
        while temp:
            print temp.data,
            temp = temp.next
            
        print("")

# Driver Code
if __name__ == "__main__":
    
    # Start with empty list
    llist = LinkedList()
    a = LinkedList()
    b = LinkedList()
    
    # Created linked list will be
    # 0->1->2->3->4->5 
    llist.push(5)
    llist.push(4)
    llist.push(3)
    llist.push(2)
    llist.push(1)
    llist.push(0)
    
    llist.AlternatingSplit(a, b)
    
    print "Original Linked List: ",
    llist.printList()
    
    print "Resultant Linked List 'a' : ",
    a.printList()
    
    print "Resultant Linked List 'b' : ",
    b.printList()
    
# This code is contributed by kevalshah5

Output:

Original linked List: 0 1 2 3 4 5 
Resultant Linked List 'a' : 4 2 0 
Resultant Linked List 'b' : 5 3 1

Time Complexity: O(n) where n is a number of nodes in the given linked list.

Space Complexity: O(1),The time complexity for this algorithm is O(n) as we are traversing through the linked list only once.
Please refer complete article on Alternating split of a given Singly Linked List | Set 1 for more details!