Find the middle of a given linked list

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++
Java
Python3
// 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 Codeint 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 Program for the above approachclass 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).
# Python program for the above approachclass 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 Codehead = Nonetemp = 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)
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 :
C++
C
Java
C#
Python3
Javascript
// 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 Codeint 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 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 listvoid 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 program to find middle of linked listclass 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# program to find middle of linked listusing System; class LinkedList{     // Head of linked list Node head; // Linked list nodeclass Node{    public int data;    public Node next;         public Node(int d)    {        data = d;        next = null;    }} // Function to print middle of linked listvoid 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 nodepublic void printList(){    Node tnode = head;    while (tnode != null)    {        Console.Write(tnode.data + "->");        tnode = tnode.next;    }    Console.WriteLine("NULL");} // Driver codestatic 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 program to find middle of linked list# Node classclass 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 objectclass 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 hereif __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)
<script> // JavaScript program to find// middle of linked listvar 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>
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++
C
Java
Python3
C#
Javascript
#include <bits/stdc++.h>using namespace std; // Link list nodestruct node{    int data;    struct node* next;}; // Function to get the middle of// the linked listvoid 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 listvoid printList(struct node* ptr){    while (ptr != NULL)    {        printf("%d->", ptr->data);        ptr = ptr->next;    }    printf("NULL\n");} // Driver codeint 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
#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 listvoid 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;}
class GFG{ static Node head; // Link list nodeclass 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 listvoid 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 listvoid printList(Node head){    while (head != null)    {        System.out.print(head.data + "-> ");        head = head.next;    }    System.out.println("null");} // Driver codepublic 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
# Node classclass 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 objectclass 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 hereif __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
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
<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 nodevar 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>
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. 
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