Given a singly linked list, find the middle of the linked list. For example, if the given linked list is 1->2->3->4->5 then the output should be 3.
If there are even nodes, then there would be two middle nodes, we need to print the second middle element. For example, if given linked list is 1->2->3->4->5->6 then output should be 4.
Method 1:
Traverse the whole linked list and count the no. of nodes. Now traverse the list again till count/2 and return the node at count/2.
Method 2:
Traverse linked list using two pointers. Move one pointer by one and the other pointers by two. When the fast pointer reaches the end slow pointer will reach the middle of the linked list.
Below image shows how printMiddle function works in the code :

C++
#include<bits/stdc++.h>
using namespace std;
struct Node
{
int data;
struct Node* next;
};
void printMiddle( struct Node *head)
{
struct Node *slow_ptr = head;
struct Node *fast_ptr = head;
if (head!=NULL)
{
while (fast_ptr != NULL && fast_ptr->next != NULL)
{
fast_ptr = fast_ptr->next->next;
slow_ptr = slow_ptr->next;
}
printf ( "The middle element is [%d]\n\n" , slow_ptr->data);
}
}
void push( struct Node** head_ref, int new_data)
{
struct Node* new_node = new Node;
new_node->data = new_data;
new_node->next = (*head_ref);
(*head_ref) = new_node;
}
void printList( struct Node *ptr)
{
while (ptr != NULL)
{
printf ( "%d->" , ptr->data);
ptr = ptr->next;
}
printf ( "NULL\n" );
}
int main()
{
struct Node* head = NULL;
for ( int i=5; i>0; i--)
{
push(&head, i);
printList(head);
printMiddle(head);
}
return 0;
}
|
C
#include<stdio.h>
#include<stdlib.h>
struct Node
{
int data;
struct Node* next;
};
void printMiddle( struct Node *head)
{
struct Node *slow_ptr = head;
struct Node *fast_ptr = head;
if (head!=NULL)
{
while (fast_ptr != NULL && fast_ptr->next != NULL)
{
fast_ptr = fast_ptr->next->next;
slow_ptr = slow_ptr->next;
}
printf ( "The middle element is [%d]\n\n" , slow_ptr->data);
}
}
void push( struct Node** head_ref, int new_data)
{
struct Node* new_node =
( struct Node*) malloc ( sizeof ( struct Node));
new_node->data = new_data;
new_node->next = (*head_ref);
(*head_ref) = new_node;
}
void printList( struct Node *ptr)
{
while (ptr != NULL)
{
printf ( "%d->" , ptr->data);
ptr = ptr->next;
}
printf ( "NULL\n" );
}
int main()
{
struct Node* head = NULL;
int i;
for (i=5; i>0; i--)
{
push(&head, i);
printList(head);
printMiddle(head);
}
return 0;
}
|
Java
class LinkedList
{
Node head;
class Node
{
int data;
Node next;
Node( int d)
{
data = d;
next = null ;
}
}
void printMiddle()
{
Node slow_ptr = head;
Node fast_ptr = head;
if (head != null )
{
while (fast_ptr != null && fast_ptr.next != null )
{
fast_ptr = fast_ptr.next.next;
slow_ptr = slow_ptr.next;
}
System.out.println( "The middle element is [" +
slow_ptr.data + "] \n" );
}
}
public void push( int new_data)
{
Node new_node = new Node(new_data);
new_node.next = head;
head = new_node;
}
public void printList()
{
Node tnode = head;
while (tnode != null )
{
System.out.print(tnode.data+ "->" );
tnode = tnode.next;
}
System.out.println( "NULL" );
}
public static void main(String [] args)
{
LinkedList llist = new LinkedList();
for ( int i= 5 ; i> 0 ; --i)
{
llist.push(i);
llist.printList();
llist.printMiddle();
}
}
}
|
Python3
class Node:
def __init__( self , data):
self .data = data
self . next = None
class LinkedList:
def __init__( self ):
self .head = None
def push( self , new_data):
new_node = Node(new_data)
new_node. next = self .head
self .head = new_node
def printList( self ):
node = self .head
while node:
print ( str (node.data) + "->" , end = "")
node = node. next
print ( "NULL" )
def printMiddle( self ):
slow = self .head
fast = self .head
while fast and fast. next :
slow = slow. next
fast = fast. next . next
print ( "The middle element is " , slow.data)
if __name__ = = '__main__' :
llist = LinkedList()
for i in range ( 5 , 0 , - 1 ):
llist.push(i)
llist.printList()
llist.printMiddle()
|
Output5->NULL
The middle element is [5]
4->5->NULL
The middle element is [5]
3->4->5->NULL
The middle element is [4]
2->3->4->5->NULL
The middle element is [4]
1->2->3->4->5->NULL
The middle element is [3]
Method 3:
Initialize mid element as head and initialize a counter as 0. Traverse the list from head, while traversing increment the counter and change mid to mid->next whenever the counter is odd. So the mid will move only half of the total length of the list.
Thanks to Narendra Kangralkar for suggesting this method.
C++
#include <bits/stdc++.h>
using namespace std;
struct node
{
int data;
struct node* next;
};
void printMiddle( struct node* head)
{
int count = 0;
struct node* mid = head;
while (head != NULL)
{
if (count & 1)
mid = mid->next;
++count;
head = head->next;
}
if (mid != NULL)
printf ( "The middle element is [%d]\n\n" ,
mid->data);
}
void push( struct node** head_ref, int new_data)
{
struct node* new_node = ( struct node*) malloc (
sizeof ( struct node));
new_node->data = new_data;
new_node->next = (*head_ref);
(*head_ref) = new_node;
}
void printList( struct node* ptr)
{
while (ptr != NULL)
{
printf ( "%d->" , ptr->data);
ptr = ptr->next;
}
printf ( "NULL\n" );
}
int main()
{
struct node* head = NULL;
int i;
for (i = 5; i > 0; i--)
{
push(&head, i);
printList(head);
printMiddle(head);
}
return 0;
}
|
C
#include <stdio.h>
#include <stdlib.h>
struct node {
int data;
struct node* next;
};
void printMiddle( struct node* head)
{
int count = 0;
struct node* mid = head;
while (head != NULL) {
if (count & 1)
mid = mid->next;
++count;
head = head->next;
}
if (mid != NULL)
printf ( "The middle element is [%d]\n\n" , mid->data);
}
void push( struct node** head_ref, int new_data)
{
struct node* new_node
= ( struct node*) malloc ( sizeof ( struct node));
new_node->data = new_data;
new_node->next = (*head_ref);
(*head_ref) = new_node;
}
void printList( struct node* ptr)
{
while (ptr != NULL) {
printf ( "%d->" , ptr->data);
ptr = ptr->next;
}
printf ( "NULL\n" );
}
int main()
{
struct node* head = NULL;
int i;
for (i = 5; i > 0; i--) {
push(&head, i);
printList(head);
printMiddle(head);
}
return 0;
}
|
Java
class GFG{
static Node head;
class Node
{
int data;
Node next;
public Node(Node next, int data)
{
this .data = data;
this .next = next;
}
}
void printMiddle(Node head)
{
int count = 0 ;
Node mid = head;
while (head != null )
{
if ((count % 2 ) == 1 )
mid = mid.next;
++count;
head = head.next;
}
if (mid != null )
System.out.println( "The middle element is [" +
mid.data + "]\n" );
}
void push(Node head_ref, int new_data)
{
Node new_node = new Node(head_ref, new_data);
head = new_node;
}
void printList(Node head)
{
while (head != null )
{
System.out.print(head.data + "-> " );
head = head.next;
}
System.out.println( "null" );
}
public static void main(String[] args)
{
GFG ll = new GFG();
for ( int i = 5 ; i > 0 ; i--)
{
ll.push(head, i);
ll.printList(head);
ll.printMiddle(head);
}
}
}
|
Python3
class Node:
def __init__( self , data):
self .data = data
self . next = None
class LinkedList:
def __init__( self ):
self .head = None
def push( self , new_data):
new_node = Node(new_data)
new_node. next = self .head
self .head = new_node
def printList( self ):
node = self .head
while node:
print ( str (node.data) + "->" , end = "")
node = node. next
print ( "NULL" )
def printMiddle( self ):
count = 0
mid = self .head
heads = self .head
while (heads ! = None ):
if count& 1 :
mid = mid. next
count + = 1
heads = heads. next
if mid! = None :
print ( "The middle element is " , mid.data)
if __name__ = = '__main__' :
llist = LinkedList()
for i in range ( 5 , 0 , - 1 ):
llist.push(i)
llist.printList()
llist.printMiddle()
|
Output5->NULL
The middle element is [5]
4->5->NULL
The middle element is [5]
3->4->5->NULL
The middle element is [4]
2->3->4->5->NULL
The middle element is [4]
1->2->3->4->5->NULL
The middle element is [3]
https://www.youtube.com/watch?v=BrmGrIPGbgk
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