Skip to content
Related Articles
Open in App
Not now

Related Articles

Recursive approach for alternating split of Linked List

Improve Article
Save Article
  • Difficulty Level : Easy
  • Last Updated : 23 Jan, 2023
Improve Article
Save Article

Given a linked list, split the linked list into two with alternate nodes.

Examples: 

Input : 1 2 3 4 5 6 7
Output : 1 3 5 7
         2 4 6

Input : 1 4 5 6
Output : 1 5
         4 6

We have discussed Iterative splitting of linked list.

The idea is to begin from two nodes first and second. Let us call these nodes as ‘a’ and ‘b’. We recurs

Implementation:

C++




// CPP code to split linked list
#include <bits/stdc++.h>
using namespace std;
 
// Node structure
struct Node {
    int data;
    struct Node* next;
};
 
// Function to push nodes
// into linked list
void push(Node** head, int new_data)
{
    Node* new_node = new Node;
    new_node->data = new_data;
    new_node->next = (*head);
    (*head) = new_node;
}
 
// We basically remove link between 'a'
// and its next. Similarly we remove link
// between 'b' and its next. Then we recur
// for remaining lists.
void moveNode(Node* a, Node* b)
{
    if (b == NULL || a == NULL)
        return;
 
    if (a->next != NULL)
        a->next = a->next->next;
 
    if (b->next != NULL)
        b->next = b->next->next;
 
    moveNode(a->next, b->next);
}
 
// function to split linked list
void alternateSplitLinkedList(Node* head, Node** aRef,
                                        Node** bRef)
{
    Node* curr = head;
    *aRef = curr;
    *bRef = curr->next;
    moveNode(*aRef, *bRef);
}
 
void display(Node* head)
{
    Node* curr = head;
    while (curr != NULL) {
        printf("%d ", curr->data);
        curr = curr->next;
    }
}
 
// Driver code
int main()
{
    Node* head = NULL;
    Node *a = NULL, *b = NULL;
 
    push(&head, 7);
    push(&head, 6);
    push(&head, 5);
    push(&head, 4);
    push(&head, 3);
    push(&head, 2);
    push(&head, 1);
 
 
    alternateSplitLinkedList(head, &a, &b);
 
    printf("a : ");
    display(a);
    printf("\nb : ");
    display(b);
 
    return 0;
}

Java




// Java code to split linked list
class GFG
{
     
// Node structure
static class Node
{
    int data;
    Node next;
};
 
// Function to push nodes
// into linked list
static Node push(Node head, int new_data)
{
    Node new_node = new Node();
    new_node.data = new_data;
    new_node.next = (head);
    (head) = new_node;
    return head;
}
 
static Node a = null, b = null;
 
// We basically remove link between 'a'
// and its next. Similarly we remove link
// between 'b' and its next. Then we recur
// for remaining lists.
static void moveNode(Node a, Node b)
{
    if (b == null || a == null)
        return;
 
    if (a.next != null)
        a.next = a.next.next;
 
    if (b.next != null)
        b.next = b.next.next;
 
    moveNode(a.next, b.next);
}
 
// function to split linked list
static void alternateSplitLinkedList(Node head)
{
    Node curr = head;
    a = curr;
    b = curr.next;
    Node aRef = a, bRef = b;
    moveNode(aRef, bRef);
}
 
static void display(Node head)
{
    Node curr = head;
    while (curr != null)
    {
        System.out.printf("%d ", curr.data);
        curr = curr.next;
    }
}
 
// Driver code
public static void main(String args[])
{
    Node head = null;
    head = push(head, 7);
    head = push(head, 6);
    head = push(head, 5);
    head = push(head, 4);
    head = push(head, 3);
    head = push(head, 2);
    head = push(head, 1);
     
    alternateSplitLinkedList(head);
 
    System.out.printf("a : ");
    display(a);
    System.out.printf("\nb : ");
    display(b);
}
}
 
// This code is contributed by Arnab Kundu

Python




# Python code to split linked list
 
# Node structure
class Node(object):
    def __init__(self, d):
        self.data = d
        self.next = None
 
# Function to push nodes
# into linked list
def push( head, new_data) :
 
    new_node = Node(0)
    new_node.data = new_data
    new_node.next = (head)
    (head) = new_node
    return head
 
a = None
b = None
 
# We basically remove link between 'a'
# and its next. Similarly we remove link
# between 'b' and its next. Then we recur
# for remaining lists.
def moveNode( a, b) :
    if (b == None or a == None) :
        return
    if (a.next != None) :
        a.next = a.next.next
    if (b.next != None) :
        b.next = b.next.next
 
    moveNode(a.next, b.next)
 
# function to split linked list
def alternateSplitLinkedList(head) :
     
    curr = head
    global a
    global b
    a = curr
    b = curr.next
    aRef = a
    bRef = b
    moveNode(aRef, bRef)
    return head;
 
def display(head) :
 
    curr = head
    while (curr != None) :
     
        print( curr.data,end = " ")
        curr = curr.next
     
# Driver code
head = None
head = push(head, 7)
head = push(head, 6)
head = push(head, 5)
head = push(head, 4)
head = push(head, 3)
head = push(head, 2)
head = push(head, 1)
     
head=alternateSplitLinkedList(head)
 
print("a : ",end="")
display(a)
print("\nb : ",end="")
display(b)
 
# This code is contributed by Arnab Kundu

C#




// C# code to split linked list
using System;
 
class GFG
{
     
// Node structure
public class Node
{
    public int data;
    public Node next;
};
 
// Function to push nodes
// into linked list
static Node push(Node head, int new_data)
{
    Node new_node = new Node();
    new_node.data = new_data;
    new_node.next = (head);
    (head) = new_node;
    return head;
}
 
static Node a = null, b = null;
 
// We basically remove link between 'a'
// and its next. Similarly we remove link
// between 'b' and its next. Then we recur
// for remaining lists.
static void moveNode(Node a, Node b)
{
    if (b == null || a == null)
        return;
 
    if (a.next != null)
        a.next = a.next.next;
 
    if (b.next != null)
        b.next = b.next.next;
 
    moveNode(a.next, b.next);
}
 
// function to split linked list
static void alternateSplitLinkedList(Node head)
{
    Node curr = head;
    a = curr;
    b = curr.next;
    Node aRef = a, bRef = b;
    moveNode(aRef, bRef);
}
 
static void display(Node head)
{
    Node curr = head;
    while (curr != null)
    {
        Console.Write("{0} ", curr.data);
        curr = curr.next;
    }
}
 
// Driver code
public static void Main(String []args)
{
    Node head = null;
    head = push(head, 7);
    head = push(head, 6);
    head = push(head, 5);
    head = push(head, 4);
    head = push(head, 3);
    head = push(head, 2);
    head = push(head, 1);
     
    alternateSplitLinkedList(head);
 
    Console.Write("a : ");
    display(a);
    Console.Write("\nb : ");
    display(b);
}
}
 
// This code is contributed by PrinciRaj1992

Javascript




<script>
 
// JavaScript code to split linked list
 
// Node structure
class Node
{
    constructor()
    {
        this.data = 0;
        this.next = null
    }
};
 
// Function to push nodes
// into linked list
function push(head, new_data)
{
    var new_node = new Node();
    new_node.data = new_data;
    new_node.next = (head);
    (head) = new_node;
    return head;
}
 
var a = null, b = null;
 
// We basically remove link between 'a'
// and its next. Similarly we remove link
// between 'b' and its next. Then we recur
// for remaining lists.
function moveNode(a, b)
{
    if (b == null || a == null)
        return;
 
    if (a.next != null)
        a.next = a.next.next;
 
    if (b.next != null)
        b.next = b.next.next;
 
    moveNode(a.next, b.next);
}
 
// function to split linked list
function alternateSplitLinkedList(head)
{
    var curr = head;
    a = curr;
    b = curr.next;
    var aRef = a, bRef = b;
    moveNode(aRef, bRef);
}
 
function display(head)
{
    var curr = head;
    while (curr != null)
    {
        document.write(curr.data+ " ");
        curr = curr.next;
    }
}
 
// Driver code
var head = null;
head = push(head, 7);
head = push(head, 6);
head = push(head, 5);
head = push(head, 4);
head = push(head, 3);
head = push(head, 2);
head = push(head, 1);
 
alternateSplitLinkedList(head);
document.write("a : ");
display(a);
document.write("<br>b : ");
display(b);
 
</script>

Output: 

a : 1 3 5 7 
b : 2 4 6 

 

Time Complexity: O(N)
Auxiliary Space: O(N)


My Personal Notes arrow_drop_up
Related Articles

Start Your Coding Journey Now!