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Java Program To Delete Middle Of Linked List

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  • Last Updated : 30 Dec, 2021

Given a singly linked list, delete the middle of the linked list. For example, if the given linked list is 1->2->3->4->5 then the linked list should be modified to 1->2->4->5

If there are even nodes, then there would be two middle nodes, we need to delete the second middle element. For example, if given linked list is 1->2->3->4->5->6 then it should be modified to 1->2->3->5->6.
If the input linked list is NULL, then it should remain NULL.

If the input linked list has 1 node, then this node should be deleted and a new head should be returned. 

Simple solution: The idea is to first count the number of nodes in a linked list, then delete n/2’th node using the simple deletion process. 

Java




// Java program to delete middle
// of a linked list
import java.io.*;
class GFG {
  
    // Link list Node 
    static class Node 
    {
        int data;
        Node next;
    }
  
    // Utility function to create 
    // a new node.
    static Node newNode(int data)
    {
        Node temp = new Node();
        temp.data = data;
        temp.next = null;
        return temp;
    }
  
    // Count of nodes
    static int countOfNodes(Node head)
    {
        int count = 0;
        while (head != null
        {
            head = head.next;
            count++;
        }
        return count;
    }
  
    // Deletes middle node and returns
    // head of the modified list
    static Node deleteMid(Node head)
    {
        // Base cases
        if (head == null)
            return null;
        if (head.next == null
        {
            return null;
        }
        Node copyHead = head;
  
        // Find the count of nodes
        int count = countOfNodes(head);
  
        // Find the middle node
        int mid = count / 2;
  
        // Delete the middle node
        while (mid-- > 1
        {
            head = head.next;
        }
  
        // Delete the middle node
        head.next = head.next.next;
  
        return copyHead;
    }
  
    // A utility function to print
    // a given linked list
    static void printList(Node ptr)
    {
        while (ptr != null
        {
            System.out.print(ptr.data + 
                             "->");
            ptr = ptr.next;
        }
        System.out.println("NULL");
    }
  
    // Driver code
    public static void main(String[] args)
    {
        // Start with the empty list 
        Node head = newNode(1);
        head.next = newNode(2);
        head.next.next = newNode(3);
        head.next.next.next = newNode(4);
  
        System.out.println("Given Linked List");
        printList(head);
        head = deleteMid(head);
        System.out.println(
               "Linked List after deletion of middle");
        printList(head);
    }
}
// This code is contributed by rajsanghavi9

Output:

Given Linked List
1->2->3->4->NULL
Linked List after deletion of middle
1->2->4->NULL

Complexity Analysis: 

  • Time Complexity: O(n). 
    Two traversals of the linked list is needed
  • Auxiliary Space: O(1). 
    No extra space is needed.

Efficient solution: 
Approach: The above solution requires two traversals of the linked list. The middle node can be deleted using one traversal. The idea is to use two pointers, slow_ptr, and fast_ptr. Both pointers start from the head of list. When fast_ptr reaches the end, slow_ptr reaches middle. This idea is same as the one used in method 2 of this post. The additional thing in this post is to keep track of the previous middle so the middle node can be deleted.

Below is the implementation.  

Java




// Java program to delete the 
// middle of a linked list
class GfG 
{
    // Link list Node 
    static class Node 
    {
        int data;
        Node next;
    }
  
    // Deletes middle node and returns
    // head of the modified list
    static Node deleteMid(Node head)
    {
        // Base cases
        if (head == null)
            return null;
        if (head.next == null
        {
            return null;
        }
  
        // Initialize slow and fast pointers 
        // to reach middle of linked list
        Node slow_ptr = head;
        Node fast_ptr = head;
  
        // Find the middle and previous 
        // of middle.
        Node prev = null;
  
        // To store previous of slow_ptr
        while (fast_ptr != null && 
               fast_ptr.next != null
        {
            fast_ptr = fast_ptr.next.next;
            prev = slow_ptr;
            slow_ptr = slow_ptr.next;
        }
  
        // Delete the middle node
        prev.next = slow_ptr.next;
  
        return head;
    }
  
    // A utility function to print 
    // a given linked list
    static void printList(Node ptr)
    {
        while (ptr != null
        {
            System.out.print(ptr.data + "->");
            ptr = ptr.next;
        }
        System.out.println("NULL");
    }
  
    // Utility function to create a 
    // new node.
    static Node newNode(int data)
    {
        Node temp = new Node();
        temp.data = data;
        temp.next = null;
        return temp;
    }
  
    // Driver code
    public static void main(String[] args)
    {
        // Start with the empty list 
        Node head = newNode(1);
        head.next = newNode(2);
        head.next.next = newNode(3);
        head.next.next.next = newNode(4);
  
        System.out.println("Given Linked List");
        printList(head);
        head = deleteMid(head);
        System.out.println("Linked List after deletion of middle");
        printList(head);
    }
}
// This code is contributed by Prerna saini

Output:

Given Linked List
1->2->3->4->NULL
Linked List after deletion of middle
1->2->4->NULL

Complexity Analysis: 

  • Time Complexity: O(n). 
    Only one traversal of the linked list is needed
  • Auxiliary Space: O(1). 
    As no extra space is needed.

Please refer complete article on Delete middle of linked list for more details!


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