Recursive approach for alternating split of Linked List
Last Updated :
23 Jan, 2023
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)