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Java Program For Finding The Middle Element Of A Given Linked List

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  • Last Updated : 22 Jun, 2022
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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 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. 

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 :

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

 

Java




// Java program to find middle of
// the linked list
class LinkedList
{
    // Head of linked list
    Node head;
 
    // Linked list node
    class Node
    {
        int data;
        Node next;
        Node(int d)
        {
            data = d;
            next = null;
        }
    }
 
    // Function to print middle of
    // the 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;
            }
            System.out.println("The middle element is [" +
                                slow_ptr.data + "]");
        }
    }
 
    // 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");
    }
  
    // Driver code
    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


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) where n is the number of nodes in the given linked list.
Auxiliary Space: O(1), no extra space is required, so it is a constant.

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.  

Java




// Java program to implement the
// above approach
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


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) where n is the number of nodes in the given linked list.
Auxiliary Space: O(1), no extra space is required, so it is a constant.

Please refer complete article on Find the middle of a given linked list for more details!


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