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  • Difficulty Level : Easy
  • Last Updated : 22 Jun, 2022

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 the given linked list is 1->2->3->4->5->6 then the 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. 
Below is the implementation of the above approach:

C++




// C++ program for the above approach
 
#include <iostream>
using namespace std;
 
class Node {
public:
    int data;
    Node* next;
};
 
class NodeOperation {
public:
    // Function to add a new node
    void pushNode(class Node** head_ref, int data_val)
    {
 
        // Allocate node
        class Node* new_node = new Node();
 
        // Put in the data
        new_node->data = data_val;
 
        // Link the old list off the new node
        new_node->next = *head_ref;
 
        // move the head to point to the new node
        *head_ref = new_node;
    }
 
    // A utility function to print a given linked list
    void printNode(class Node* head)
    {
        while (head != NULL) {
            cout << head->data << "->";
            head = head->next;
        }
        cout << "NULL" << endl;
    }
 
    /* Utility Function to find length of linked list */
    int getLen(class Node* head)
    {
        int len = 0;
        class Node* temp = head;
        while (temp) {
            len++;
            temp = temp->next;
        }
        return len;
    }
 
    void printMiddle(class Node* head)
    {
 
        if (head) {
            // find length
            int len = getLen(head);
            class Node* temp = head;
 
            // trvaers till we reached half of length
            int midIdx = len / 2;
            while (midIdx--) {
                temp = temp->next;
            }
            // temp will be storing middle element
            cout << "The middle element is [" << temp->data
                 << "]" << endl;
        }
    }
};
 
// Driver Code
int main()
{
    class Node* head = NULL;
    class NodeOperation* temp = new NodeOperation();
    for (int i = 5; i > 0; i--) {
        temp->pushNode(&head, i);
        temp->printNode(head);
        temp->printMiddle(head);
    }
    return 0;
}
 
// This code is contributed by Tapesh(tapeshdua420)


Java




// Java Program for the above approach
class GFG {
   
      Node head;
 
    /*Creating a new Node*/
    class Node {
        int data;
        Node next;
        public Node(int data)
        {
            this.data = data;
            this.next = null;
        }
    }
 
    /*Function to add a new Node*/
    public void pushNode(int data)
    {
        Node new_node = new Node(data);
          new_node.next = head;
          head = new_node;
    }
 
    /*Displaying the elements in the list*/
    public void printNode()
    {
        Node temp = head;
        while (temp != null) {
            System.out.print(temp.data + "->");
            temp = temp.next;
        }
        System.out.print("Null"+"\n");
    }
 
    /*Finding the length of the list.*/
    public int getLen()
    {
        int length = 0;
        Node temp = head;
        while (temp != null) {
            length++;
            temp = temp.next;
        }
        return length;
    }
 
    /*Printing the middle element of the list.*/
    public void printMiddle()
    {
        if (head != null) {
            int length = getLen();
            Node temp = head;
            int middleLength = length / 2;
            while (middleLength != 0) {
                temp = temp.next;
                middleLength--;
            }
            System.out.print("The middle element is ["
                             + temp.data + "]");
            System.out.println();
        }
    }
 
    public static void main(String[] args)
    {
        GFG list = new GFG();
        for (int i = 5; i >= 1; i--) {
            list.pushNode(i);
            list.printNode();
            list.printMiddle();
        }
    }
}
 
// This Code is contributed by lokesh (lokeshmvs21).


Python3




# Python program for the above approach
class Node:
    def __init__(self, data):
        self.data = data
        self.next = None
 
 
class NodeOperation:
    # Function to add a new node
    def pushNode(self, head_ref, data_val):
 
        # Allocate node and put in the data
        new_node = Node(data_val)
 
        # Link the old list off the new node
        new_node.next = head_ref
 
        # move the head to point to the new node
        head_ref = new_node
        return head_ref
 
    # A utility function to print a given linked list
    def printNode(self, head):
        while (head != None):
            print('%d->' % head.data, end="")
            head = head.next
        print("NULL")
 
    ''' Utility Function to find length of linked list '''
 
    def getLen(self, head):
        temp = head
        len = 0
 
        while (temp != None):
            len += 1
            temp = temp.next
 
        return len
 
    def printMiddle(self, head):
        if head != None:
            # find length
            len = self.getLen(head)
            temp = head
 
            # traverse till we reached half of length
            midIdx = len // 2
            while midIdx != 0:
                temp = temp.next
                midIdx -= 1
 
            # temp will be storing middle element
            print('The middle element is [%d]' % temp.data)
 
 
# Driver Code
head = None
temp = NodeOperation()
for i in range(5, 0, -1):
    head = temp.pushNode(head, i)
    temp.printNode(head)
    temp.printMiddle(head)
 
# This code is contributed by Tapesh(tapeshdua420)


Output

5->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 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, the slow pointer will reach the middle of the linked list.

Below image shows how printMiddle function works in the code :

middle-of-a-given-linked-list-in-C-and-Java1

C++




// C++ program for the above approach
 
#include <iostream>
using namespace std;
 
class Node{
    public:
        int data;
        Node *next;
};
 
class NodeOperation{
  public:
   
    // Function to add a new node
    void pushNode(class Node** head_ref,int data_val){
       
        // Allocate node
        class Node *new_node = new Node();
         
         // Put in the data
        new_node->data = data_val;
         
        // Link the old list off the new node
        new_node->next = *head_ref;
         
        // move the head to point to the new node
        *head_ref = new_node;
    }
     
    // A utility function to print a given linked list
    void printNode(class Node *head){
        while(head != NULL){
            cout <<head->data << "->";
            head = head->next;
        }
        cout << "NULL" << endl;
    }
     
    void printMiddle(class 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;
            }
            cout << "The middle element is [" << slow_ptr->data << "]" << endl;
        }
    }
};
 
// Driver Code
int main(){
    class Node *head = NULL;
    class NodeOperation *temp = new NodeOperation();
    for(int i=5; i>0; i--){
        temp->pushNode(&head, i);
        temp->printNode(head);
        temp->printMiddle(head);
    }
    return 0;
}


C




// C program to find middle of linked list
#include<stdio.h>
#include<stdlib.h>
 
/* Link list node */
struct Node
{
    int data;
    struct Node* next;
};
 
/* Function to get the middle of the linked list*/
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)
{
    /* 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 off the new node */
    new_node->next = (*head_ref);
 
    /* move the head to point to the new node */
    (*head_ref) = new_node;
}
 
// A utility function to print a given linked list
void printList(struct Node *ptr)
{
    while (ptr != NULL)
    {
        printf("%d->", ptr->data);
        ptr = ptr->next;
    }
    printf("NULL\n");
}
 
/* Driver program to test above function*/
int main()
{
    /* Start with the empty list */
    struct Node* head = NULL;
    int i;
 
    for (i=5; i>0; i--)
    {
        push(&head, i);
        printList(head);
        printMiddle(head);
    }
 
    return 0;
}


Java




// Java program to find middle of linked list
class LinkedList
{
    Node head; // head of linked list
 
    /* Linked list node */
    class Node
    {
        int data;
        Node next;
        Node(int d)
        {
            data = d;
            next = null;
        }
    }
 
    /* Function to print middle of linked list */
    void printMiddle()
    {
        Node slow_ptr = head;
        Node fast_ptr = head;
         
            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");
         
    }
 
    /* Inserts a new Node at front of the list. */
    public void push(int new_data)
    {
        /* 1 & 2: Allocate the Node &
                  Put in the data*/
        Node new_node = new Node(new_data);
 
        /* 3. Make next of new Node as head */
        new_node.next = head;
 
        /* 4. Move the head to point to new Node */
        head = new_node;
    }
 
    /* This function prints contents of linked list
       starting from  the given 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();
        }
    }
}
// This code is contributed by Rajat Mishra


C#




// C# program to find middle of linked list
using System;
 
class LinkedList{
     
// Head of linked list 
Node head;
 
// Linked list node
class Node
{
    public int data;
    public Node next;
     
    public Node(int d)
    {
        data = d;
        next = null;
    }
}
 
// Function to print middle of linked list
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;
        }
        Console.WriteLine("The middle element is [" +
                          slow_ptr.data + "] \n");
    }
}
 
// Inserts a new Node at front of the list.
public void push(int new_data)
{
     
    /* 1 & 2: Allocate the Node &
              Put in the data*/
    Node new_node = new Node(new_data);
 
    /* 3. Make next of new Node as head */
    new_node.next = head;
 
    /* 4. Move the head to point to new Node */
    head = new_node;
}
 
// This function prints contents of linked
// list starting from  the given node
public void printList()
{
    Node tnode = head;
    while (tnode != null)
    {
        Console.Write(tnode.data + "->");
        tnode = tnode.next;
    }
    Console.WriteLine("NULL");
}
 
// Driver code
static public void Main()
{
    LinkedList llist = new LinkedList();
    for(int i = 5; i > 0; --i)
    {
        llist.push(i);
        llist.printList();
        llist.printMiddle();
    }
}
}
 
// This code is contributed by Dharanendra L V.


Python3




# Python3 program to find middle of linked list
# Node class
class Node:
   
    # Function to initialise the node object
    def __init__(self, data):
        self.data = data  # Assign data
        self.next = None  # Initialize next as null
   
   
# Linked List class contains a Node object
class LinkedList:
   
    # Function to initialize head
    def __init__(self):
        self.head = None
 
    # Function to insert a new node at the beginning 
    def push(self, new_data): 
        new_node = Node(new_data) 
        new_node.next = self.head 
        self.head = new_node
 
    # Print the linked list
    def printList(self):
        node = self.head
        while node:
            print(str(node.data) + "->", end="")
            node = node.next
        print("NULL")
 
    # Function that returns middle.
    def printMiddle(self):
        # Initialize two pointers, one will go one step a time (slow), another two at a time (fast)
        slow = self.head
        fast = self.head
 
        # Iterate till fast's next is null (fast reaches end)
        while fast and fast.next:
            slow = slow.next
            fast = fast.next.next
         
        # return the slow's data, which would be the middle element.
        print("The middle element is ", slow.data)
 
# Code execution starts here
if __name__=='__main__':
   
    # Start with the empty list
    llist = LinkedList()
   
    for i in range(5, 0, -1):
        llist.push(i)
        llist.printList()
        llist.printMiddle()
 
 # Code is contributed by Kumar Shivam (kshivi99)


Javascript




<script>
 
// JavaScript program to find
// middle of linked list
var head; // head of linked list
 
    /* Linked list node */
    class Node {
        constructor(val) {
            this.data = val;
            this.next = null;
        }
    }
 
    /* Function to print middle
    of linked list */
    function printMiddle()
    {
   var slow_ptr = head;
   var 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;
            }
            document.write(
"The middle element is [" + slow_ptr.data + "] <br/><br/>"
            );
        }
    }
 
    /* Inserts a new Node at front of the list. */
     function push(new_data) {
        /*
         * 1 & 2: Allocate the Node & Put in the data
         */
var new_node = new Node(new_data);
 
        /* 3. Make next of new Node as head */
        new_node.next = head;
 
        /* 4. Move the head to point to new Node */
        head = new_node;
    }
 
    /*
     * This function prints contents of linked
     list starting from the given node
     */
     function printList() {
     var tnode = head;
        while (tnode != null) {
            document.write(tnode.data + "->");
            tnode = tnode.next;
        }
        document.write("NULL<br/>");
    }
 
     
        for (i = 5; i > 0; --i) {
            push(i);
            printList();
            printMiddle();
        }
 
// This code is contributed by todaysgaurav
 
</script>


Output

5->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]

Time Complexity: O(N), As we are traversing the list only once.
Auxiliary Space: O(1), As constant extra space is used.

Method 3: Initialize the mid element as head and initialize a counter as 0. Traverse the list from the 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;
 
// Link list node
struct node
{
    int data;
    struct node* next;
};
 
// Function to get the middle of
// the linked list
void printMiddle(struct node* head)
{
    int count = 0;
    struct node* mid = head;
 
    while (head != NULL)
    {
         
        // Update mid, when 'count'
        // is odd number
        if (count & 1)
            mid = mid->next;
 
        ++count;
        head = head->next;
    }
 
    // If empty list is provided
    if (mid != NULL)
        printf("The middle element is [%d]\n\n",
                mid->data);
}
 
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 off the new node
    new_node->next = (*head_ref);
 
    // Move the head to point to
    // the new node
    (*head_ref) = new_node;
}
 
// A utility function to print
// a given linked list
void printList(struct node* ptr)
{
    while (ptr != NULL)
    {
        printf("%d->", ptr->data);
        ptr = ptr->next;
    }
    printf("NULL\n");
}
 
// Driver code
int main()
{
     
    // Start with the empty list
    struct node* head = NULL;
    int i;
 
    for(i = 5; i > 0; i--)
    {
        push(&head, i);
        printList(head);
        printMiddle(head);
    }
    return 0;
}
 
// This code is contributed by ac121102


C




#include <stdio.h>
#include <stdlib.h>
 
/* Link list node */
struct node {
    int data;
    struct node* next;
};
 
/* Function to get the middle of the linked list*/
void printMiddle(struct node* head)
{
    int count = 0;
    struct node* mid = head;
 
    while (head != NULL) {
        /* update mid, when 'count' is odd number */
        if (count & 1)
            mid = mid->next;
 
        ++count;
        head = head->next;
    }
 
    /* if empty list is provided */
    if (mid != NULL)
        printf("The middle element is [%d]\n\n", mid->data);
}
 
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 off the new node */
    new_node->next = (*head_ref);
 
    /* move the head to point to the new node */
    (*head_ref) = new_node;
}
 
// A utility function to print a given linked list
void printList(struct node* ptr)
{
    while (ptr != NULL) {
        printf("%d->", ptr->data);
        ptr = ptr->next;
    }
    printf("NULL\n");
}
 
/* Driver program to test above function*/
int main()
{
    /* Start with the empty list */
    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;
 
// Link list node
class Node
{
    int data;
    Node next;
 
    // Constructor
    public Node(Node next, int data)
    {
        this.data = data;
        this.next = next;
    }
}
 
// Function to get the middle of
// the linked list
void printMiddle(Node head)
{
    int count = 0;
    Node mid = head;
 
    while (head != null)
    {
 
        // Update mid, when 'count'
        // is odd number
        if ((count % 2) == 1)
            mid = mid.next;
 
        ++count;
        head = head.next;
    }
 
    // If empty list is provided
    if (mid != null)
        System.out.println("The middle element is [" +
                            mid.data + "]\n");
}
 
void push(Node head_ref, int new_data)
{
     
    // Allocate node
    Node new_node = new Node(head_ref, new_data);
 
    // Move the head to point to the new node
    head = new_node;
}
 
// A utility function to print a
// given linked list
void printList(Node head)
{
    while (head != null)
    {
        System.out.print(head.data + "-> ");
        head = head.next;
    }
    System.out.println("null");
}
 
// Driver code
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);
    }
}
}
 
// This code is contributed by mark_3


Python3




# Node class
class Node:
    
    # Function to initialise the node object
    def __init__(self, data):
        self.data = data  # Assign data
        self.next = None  # Initialize next as null
    
# Linked List class contains a Node object
class LinkedList:
    
    # Function to initialize head
    def __init__(self):
        self.head = None
  
    # Function to insert a new node at the beginning 
    def push(self, new_data): 
        new_node = Node(new_data) 
        new_node.next = self.head 
        self.head = new_node
  
    # Print the linked list
    def printList(self):
        node = self.head
        while node:
            print(str(node.data) + "->", end = "")
            node = node.next
        print("NULL")
  
    # Function to get the middle of
    #  the linked list
    def printMiddle(self):
        count = 0
        mid = self.head
        heads = self.head
   
        while(heads != None):
       
        # Update mid, when 'count'
        # is odd number
            if count&1:
                mid = mid.next
            count += 1
            heads = heads.next
             
        # If empty list is provided
        if mid!=None:
            print("The middle element is ", mid.data)
  
# Code execution starts here
if __name__=='__main__':
    
    # Start with the empty list
    llist = LinkedList()
    
    for i in range(5, 0, -1):
        llist.push(i)
        llist.printList()
        llist.printMiddle()
  
 # This Code is contributed by Manisha_Ediga


C#




using System;
 
public class GFG
{
 
    static Node head;
 
    // Link list node
    public
 
 class Node {
        public
 
 int data;
        public
 
 Node next;
 
        // Constructor
        public Node(Node next, int data) {
            this.data = data;
            this.next = next;
        }
    }
 
    // Function to get the middle of
    // the linked list
    void printMiddle(Node head) {
        int count = 0;
        Node mid = head;
 
        while (head != null) {
 
            // Update mid, when 'count'
            // is odd number
            if ((count % 2) == 1)
                mid = mid.next;
 
            ++count;
            head = head.next;
        }
 
        // If empty list is provided
        if (mid != null)
            Console.WriteLine("The middle element is [" + mid.data + "]\n");
    }
 
    public void Push(Node head_ref, int new_data) {
 
        // Allocate node
        Node new_node = new Node(head_ref, new_data);
 
        // Move the head to point to the new node
        head = new_node;
    }
 
    // A utility function to print a
    // given linked list
    void printList(Node head) {
        while (head != null) {
            Console.Write(head.data + "-> ");
            head = head.next;
        }
        Console.WriteLine("null");
    }
 
    // Driver code
    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);
        }
    }
}
 
 
// This code is contributed by Rajput-Ji


Javascript




<script>
    var head=null;
 
    // Link list node
     class Node {
            constructor(next,val) {
                this.data = val;
                this.next = next;
            }
        }
 
    // Function to get the middle of
    // the linked list
    function printMiddle(head) {
        var count = 0;
var mid = head;
 
        while (head != null) {
 
            // Update mid, when 'count'
            // is odd number
            if ((count % 2) == 1)
                mid = mid.next;
 
            ++count;
            head = head.next;
        }
 
        // If empty list is provided
        if (mid != null)
            document.write("The middle element is [" + mid.data + "]<br/><br/>");
    }
 
    function push(head_ref , new_data) {
 
        // Allocate node
var new_node = new Node(head_ref, new_data);
 
        // Move the head to point to the new node
        head = new_node;
        return head;
    }
 
    // A utility function to print a
    // given linked list
    function printList(head) {
        while (head != null) {
            document.write(head.data + "-> ");
            head = head.next;
        }
        document.write("null<br/>");
    }
 
    // Driver code
     
        for (i = 5; i > 0; i--) {
        head=    push(head, i);
            printList(head);
            printMiddle(head);
        }
 
// This code contributed by gauravrajput1
</script>


Output

5->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]

Time Complexity: O(N), As we are traversing the list once.
Auxiliary Space: O(1), As constant extra space is used.
 

 Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above. 

 


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