Alternating split of a given Singly Linked List | Set 1

Last Updated : 14 Mar, 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 1(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.

C++

/* C++ Program to alternatively split
a linked list into two halves */
#include <bits/stdc++.h>
using namespace std;
 
/* Link list node */
class Node 
{ 
    public:
    int data; 
    Node* next; 
}; 
 
/* pull off the front node of
the source and put it in dest */
void MoveNode(Node** destRef, Node** sourceRef) ; 
 
/* 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. */
void AlternatingSplit(Node* source, Node** aRef, 
                            Node** bRef) 
{ 
    /* split the nodes of source 
    to these 'a' and 'b' lists */
    Node* a = NULL; 
    Node* b = NULL; 
         
    Node* current = source; 
    while (current != NULL) 
    { 
        MoveNode(&a, ¤t); /* Move a node to list 'a' */
        if (current != NULL) 
        { 
            MoveNode(&b, ¤t); /* Move a node to list 'b' */
        } 
    } 
    *aRef = a; 
    *bRef = b; 
} 
 
/* Take the node from the front of
the source, and move it to the front
of the dest. It is an error to call
this with the source list empty. 
     
Before calling MoveNode(): 
source == {1, 2, 3} 
dest == {1, 2, 3} 
         
After calling MoveNode(): 
source == {2, 3}     
dest == {1, 1, 2, 3}     
*/
void MoveNode(Node** destRef, Node** sourceRef) 
{ 
    /* the front source node */
    Node* newNode = *sourceRef; 
    assert(newNode != NULL); 
         
    /* Advance the source pointer */
    *sourceRef = newNode->next; 
         
    /* Link the old dest of the new node */
    newNode->next = *destRef; 
         
    /* Move dest to point to the new node */
    *destRef = newNode; 
} 
 
/* UTILITY FUNCTIONS */
/* Function to insert a node at 
the beginning of the linked list */
void push(Node** head_ref, int new_data) 
{ 
    /* allocate node */
    Node* new_node = new Node();
     
    /* put in the data */
    new_node->data = new_data; 
     
    /* link the old list of the new node */
    new_node->next = (*head_ref);     
     
    /* move the head to point to the new node */
    (*head_ref) = new_node; 
} 
 
/* Function to print nodes
in a given linked list */
void printList(Node *node) 
{ 
    while(node!=NULL) 
    { 
    cout<<node->data<<" "; 
    node = node->next; 
    } 
} 
 
/* Driver code*/
int main() 
{ 
    /* Start with the empty list */
    Node* head = NULL; 
    Node* a = NULL; 
    Node* b = NULL; 
     
    /* Let us create a sorted linked list to test the functions 
    Created linked list will be 0->1->2->3->4->5 */
    push(&head, 5); 
    push(&head, 4); 
    push(&head, 3); 
    push(&head, 2); 
    push(&head, 1);                                 
    push(&head, 0); 
     
    cout<<"Original linked List: "; 
    printList(head); 
     
    /* Remove duplicates from linked list */
    AlternatingSplit(head, &a, &b); 
     
    cout<<"\nResultant Linked List 'a' : "; 
    printList(a);         
     
    cout<<"\nResultant Linked List 'b' : "; 
    printList(b);         
     
    return 0; 
} 
 
// This code is contributed by rathbhupendra

C

/*Program to alternatively split a linked list into two halves */
#include<stdio.h>
#include<stdlib.h>
#include<assert.h>
 
/* Link list node */
struct Node
{
    int data;
    struct Node* next;
};
 
/* pull off the front node of the source and put it in dest */
void MoveNode(struct Node** destRef, struct Node** sourceRef) ;
 
/* 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. */
void AlternatingSplit(struct Node* source, struct Node** aRef, 
                            struct Node** bRef) 
{
  /* split the nodes of source to these 'a' and 'b' lists */
  struct Node* a = NULL; 
  struct Node* b = NULL;
   
  struct Node* current = source;
  while (current != NULL) 
  {
    MoveNode(&a, ¤t); /* Move a node to list 'a' */
    if (current != NULL) 
    {
       MoveNode(&b, ¤t); /* Move a node to list 'b' */
    }
  }
  *aRef = a;
  *bRef = b;
}
 
/* Take the node from the front of the source, and move it to the front of the dest.
   It is an error to call this with the source list empty. 
    
   Before calling MoveNode():
   source == {1, 2, 3}   
   dest == {1, 2, 3}
       
   After calling MoveNode():
   source == {2, 3}      
   dest == {1, 1, 2, 3}      
*/
void MoveNode(struct Node** destRef, struct Node** sourceRef) 
{
  /* the front source node  */
  struct Node* newNode = *sourceRef; 
  assert(newNode != NULL);
   
  /* Advance the source pointer */
  *sourceRef = newNode->next;
   
  /* Link the old dest of the new node */
  newNode->next = *destRef; 
   
  /* Move dest to point to the new node */
  *destRef = newNode; 
}
 
/* UTILITY FUNCTIONS */
/* Function to insert a node at the beginning of the linked list */
void push(struct node** head_ref, int new_data)
{
  /* allocate node */
  struct Node* new_node =
            (struct Node*) malloc(sizeof(struct Node));
 
  /* put in the data  */
  new_node->data  = new_data;
 
  /* link the old list of the new node */
  new_node->next = (*head_ref);     
 
  /* move the head to point to the new node */
  (*head_ref)    = new_node;
}
 
/* Function to print nodes in a given linked list */
void printList(struct Node *node)
{
  while(node!=NULL)
  {
   printf("%d ", node->data);
   node = node->next;
  }
} 
 
/* Driver program to test above functions*/
int main()
{
  /* Start with the empty list */
  struct Node* head = NULL;
  struct Node* a = NULL;
  struct Node* b = NULL;  
 
  /* Let us create a sorted linked list to test the functions
   Created linked list will be 0->1->2->3->4->5 */
  push(&head, 5);
  push(&head, 4);
  push(&head, 3);
  push(&head, 2);
  push(&head, 1);                                    
  push(&head, 0);  
 
  printf("\n Original linked List:  ");
  printList(head); 
 
  /* Remove duplicates from linked list */
  AlternatingSplit(head, &a, &b); 
 
  printf("\n Resultant Linked List 'a' ");
  printList(a);            
 
  printf("\n Resultant Linked List 'b' ");
  printList(b);            
 
  getchar();
  return 0;
}

Java

import java.util.*;
 
// Linked list node
class Node {
  int data;
  Node next;
 
  Node(int data) {
    this.data = data;
    next = null;
  }
}
 
class Main
{
   
  // Given the source list, split its nodes into two shorter lists.
  // 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.
  static void AlternatingSplit(Node source, Node[] aRef, Node[] bRef) {
    Node a = null, b = null;
    Node current = source;
    int count = 0;
 
    while (current != null) {
      if (count % 2 == 0) {
        // Move a node to list 'a'
        if (a == null) {
          aRef[0] = current;
          a = current;
        } else {
          a.next = current;
          a = a.next;
        }
      } else {
        // Move a node to list 'b'
        if (b == null) {
          bRef[0] = current;
          b = current;
        } else {
          b.next = current;
          b = b.next;
        }
      }
      current = current.next;
      count++;
    }
 
    if (a != null) {
      a.next = null;
    }
 
    if (b != null) {
      b.next = null;
    }
  }
 
  // Function to print nodes in a given linked list
  static void printList(Node node) {
    while (node != null) {
      System.out.print(node.data + " ");
      node = node.next;
    }
  }
 
  // Driver code
  public static void main(String[] args)
  {
     
    // Start with the empty list
    Node head = null;
 
    // Let us create a sorted linked list to test the functions
    // Created linked list will be 0->1->2->3->4->5
    for (int i = 5; i >= 0; i--) {
      Node newNode = new Node(i);
      newNode.next = head;
      head = newNode;
    }
 
    System.out.print("Original linked List: ");
    printList(head);
 
    Node[] aRef = new Node[1];
    Node[] bRef = new Node[1];
 
    // Remove duplicates from linked list
    AlternatingSplit(head, aRef, bRef);
 
    System.out.print("\nResultant Linked List 'a': ");
    printList(aRef[0]);
 
    System.out.print("\nResultant Linked List 'b': ");
    printList(bRef[0]);
  }
}

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 of 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

C#

// C# program to alternatively split
// a linked list into two halves
using System;
using System.Collections.Generic;
 
public class Node{
  public int data;
  public Node next;
 
  public Node(int item){
    data = item;
    next = null;
  }
}
 
public class LinkedList{
  Node head;
  // 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.
  public void AlternatingSplit(LinkedList a, LinkedList b){
    Node first = head;
    Node second = first.next;
 
    while(first != null && second != null && first.next != null)
    {
       
      // move a node to list 'a'
      MoveNode(a, first);
 
      // move a node to list 'b'
      MoveNode(b, second);
 
      first = first.next.next;
      if(first == null)
        break;
 
      second = first.next;
    }
  }
 
  // Pull off the front node of the 
  // source and put it in dest
  public void MoveNode(LinkedList dest, Node node){
 
    // Make the new node
    Node new_node = new Node(node.data);
 
    if(dest.head == null)
      dest.head = new_node;
    else{
      // Link the old dest of 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 
  void push(int data){
    // 1 & 2 allocate the Node &
    // put the data
    Node new_node = new Node(data);
 
    // Make the next of new Node as head
    new_node.next = head;
 
    // Move the head to point to new Node
    head = new_node;
  }
 
  // Function to print nodes 
  // in a given linked list 
  public void printList(){
    Node temp = head;
    while(temp != null){
      Console.Write(temp.data + " ");
      temp = temp.next;
    }
    Console.WriteLine("");
  }
  public static void Main(string[] args){
    LinkedList llist = new LinkedList();
    LinkedList a = new LinkedList();
    LinkedList b = new 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);
 
    Console.WriteLine("Original Linked List : ");
    llist.printList();
 
    Console.WriteLine("Resultant Linked List 'a' : ");
    a.printList();
 
    Console.WriteLine("Resultant Linked List 'b' : ");
    b.printList();
  }
}
 
// THIS CODE IS CONTRIBUTED BY YASH AGARWAL(YASHAGAWRAL2852002)

Javascript

<script>
// JavaScript program to alternatively split 
// a linked list into two halves 
 
// Node class
class Node{
     
    constructor(data,next = null){ 
        this.data = data
        this.next = next
    }
}
 
class LinkedList
{    
    constructor()
    {    
        this.head = null
    }
     
    // 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.
    AlternatingSplit(a, b){
         
        let first = this.head
        let second = first.next
         
        while (first != null &&
              second != null &&
          first.next != null){
               
              // Move a node to list 'a'
              this.MoveNode(a, first) 
               
              // Move a node to list 'b'
              this.MoveNode(b, second) 
               
              first = first.next.next
               
              if(first == null)
                break
               
              second = first.next
        }
    }
             
    // Pull off the front node of the 
    // source and put it in dest
    MoveNode(dest, node){
         
        // Make the new node
        let new_node = new Node(node.data)
         
        if(dest.head == null)
            dest.head = new_node
        else{
             
            // Link the old dest of 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 
    push(data){
         
        // 1 & 2 allocate the Node & 
        // put the data
         
        let new_node = new Node(data)
         
        // Make the next of new Node as head
        new_node.next = this.head
         
        // Move the head to point to new Node
        this.head = new_node
    }
         
    // Function to print nodes 
    // in a given linked list 
    printList(){
         
        let temp = this.head
        while(temp){
            document.write(temp.data," ");
            temp = temp.next
        }
             
        document.write("</br>")
    }
}
 
// Driver Code
     
// Start with empty list
let llist = new LinkedList()
let a = new LinkedList()
let b = new 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)
 
document.write("Original Linked List: ");
llist.printList()
 
document.write("Resultant Linked List 'a' : ");
a.printList()
 
document.write("Resultant Linked List 'b' : ");
b.printList()
     
// This code is contributed by shinjanpatra
 
</script>

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.

Auxiliary Space: O(1)

As constant extra space is used.
Method 2(Using Dummy Nodes)
Here is an alternative approach that builds the sub-lists in the same order as the source list. The code uses temporary dummy header nodes for the ‘a’ and ‘b’ lists as they are being built. Each sublist has a “tail” pointer that points to its current last node — that way new nodes can be appended to the end of each list easily. The dummy nodes give the tail pointers something to point to initially. The dummy nodes are efficient in this case because they are temporary and allocated in the stack. Alternately, local “reference pointers” (which always point to the last pointer in the list instead of to the last node) could be used to avoid Dummy nodes.

C++

void AlternatingSplit(Node* source, 
                      Node** aRef, Node** bRef) 
{ 
    Node aDummy; 
     
    /* points to the last node in 'a' */
    Node* aTail = &aDummy; 
    Node bDummy; 
     
    /* points to the last node in 'b' */
    Node* bTail = &bDummy; 
    Node* current = source; 
    aDummy.next = NULL; 
    bDummy.next = NULL; 
    while (current != NULL) 
    { 
        MoveNode(&(aTail->next), ¤t); /* add at 'a' tail */
        aTail = aTail->next; /* advance the 'a' tail */
        if (current != NULL) 
        { 
            MoveNode(&(bTail->next), ¤t); 
            bTail = bTail->next; 
        } 
    } 
    *aRef = aDummy.next; 
    *bRef = bDummy.next; 
} 
 
// This code is contributed 
// by rathbhupendra

C

void AlternatingSplit(struct Node* source, struct Node** aRef, 
                            struct Node** bRef) 
{
  struct Node aDummy;
  struct Node* aTail = &aDummy; /* points to the last node in 'a' */
  struct Node bDummy;
  struct Node* bTail = &bDummy; /* points to the last node in 'b' */
  struct Node* current = source;
  aDummy.next = NULL;
  bDummy.next = NULL;
  while (current != NULL) 
  {
    MoveNode(&(aTail->next), ¤t); /* add at 'a' tail */
    aTail = aTail->next; /* advance the 'a' tail */
    if (current != NULL) 
    {
      MoveNode(&(bTail->next), ¤t);
      bTail = bTail->next;
    }
  }
  *aRef = aDummy.next;
  *bRef = bDummy.next;
}

Java

static void AlternatingSplit(Node source, Node aRef, 
                            Node bRef) 
{
  Node aDummy = new Node();
  Node aTail = aDummy; /* points to the last node in 'a' */
  Node bDummy = new Node();
  Node bTail = bDummy; /* points to the last node in 'b' */
  Node current = source;
  aDummy.next = null;
  bDummy.next = null;
  while (current != null) 
  {
    MoveNode((aTail.next), current); /* add at 'a' tail */
    aTail = aTail.next; /* advance the 'a' tail */
    if (current != null) 
    {
      MoveNode((bTail.next), current);
      bTail = bTail.next;
    }
  }
  aRef = aDummy.next;
  bRef = bDummy.next;
}
 
// This code is contributed by rutvik_56

Python3

def AlternatingSplit(source, aRef,  bRef):
   
  aDummy = Node();
  aTail = aDummy; ''' points to the last Node in 'a' '''
  bDummy = Node();
  bTail = bDummy; ''' points to the last Node in 'b' '''
  current = source;
  aDummy.next = None;
  bDummy.next = None;
  while (current != None):
    MoveNode((aTail.next), current); ''' add at 'a' tail '''
    aTail = aTail.next; ''' advance the 'a' tail '''
    if (current != None):
      MoveNode((bTail.next), current);
      bTail = bTail.next;
     
  aRef = aDummy.next;
  bRef = bDummy.next;
 
# This code is contributed by umadevi9616 

C#

static void AlternatingSplit(Node source, Node aRef, 
                            Node bRef) 
{
  Node aDummy = new Node();
  Node aTail = aDummy; /* points to the last node in 'a' */
  Node bDummy = new Node();
  Node bTail = bDummy; /* points to the last node in 'b' */
  Node current = source;
  aDummy.next = null;
  bDummy.next = null;
  while (current != null) 
  {
    MoveNode((aTail.next), current); /* add at 'a' tail */
    aTail = aTail.next; /* advance the 'a' tail */
    if (current != null) 
    {
      MoveNode((bTail.next), current);
      bTail = bTail.next;
    }
  }
  aRef = aDummy.next;
  bRef = bDummy.next;
}
 
// This code is contributed by pratham_76

Javascript

<script>
function AlternatingSplit( source,  aRef, 
                             bRef) 
{
  var aDummy = new Node();
  var aTail = aDummy; /* points to the last node in 'a' */
  var bDummy = new Node();
  var bTail = bDummy; /* points to the last node in 'b' */
  var current = source;
  aDummy.next = null;
  bDummy.next = null;
  while (current != null) 
  {
    MoveNode((aTail.next), current); /* add at 'a' tail */
    aTail = aTail.next; /* advance the 'a' tail */
    if (current != null) 
    {
      MoveNode((bTail.next), current);
      bTail = bTail.next;
    }
  }
  aRef = aDummy.next;
  bRef = bDummy.next;
}
 
// This code contributed by aashish1995
</script>

Time Complexity: O(n)

where n is number of node in the given linked list.

Auxiliary Space: O(1)

As Constant extra space is used.

Source: http://cslibrary.stanford.edu/105/LinkedListProblems.pdf
Please write comments if you find the above code/algorithm incorrect, or find better ways to solve the same problem.