Delete middle of linked list

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

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



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// C++ program to delete middle
// of a linked list
#include <bits/stdc++.h>
using namespace std;
  
/* Link list Node */
struct Node {
    int data;
    struct Node* next;
};
// count of nodes
int countOfNodes(struct Node* head)
{
    int count = 0;
    while (head != NULL) {
        head = head->next;
        count++;
    }
    return count;
}
  
// Deletes middle node and returns
// head of the modified list
struct Node* deleteMid(struct Node* head)
{
    // Base cases
    if (head == NULL)
        return NULL;
    if (head->next == NULL) {
        delete head;
        return NULL;
    }
    struct 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
void printList(struct Node* ptr)
{
    while (ptr != NULL) {
        cout << ptr->data << "->";
        ptr = ptr->next;
    }
    cout << "NULL\n";
}
  
// Utility function to create a new node.
Node* newNode(int data)
{
    struct Node* temp = new Node;
    temp->data = data;
    temp->next = NULL;
    return temp;
}
  
/* Drier program to test above function*/
int main()
{
    /* Start with the empty list */
    struct Node* head = newNode(1);
    head->next = newNode(2);
    head->next->next = newNode(3);
    head->next->next->next = newNode(4);
  
    cout << "Gven Linked List\n";
    printList(head);
  
    head = deleteMid(head);
  
    cout << "Linked List after deletion of middle\n";
    printList(head);
  
    return 0;
}

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Output:

Gven 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 linked list is needed
  • Auxiliary Space: O(1).
    No extra space is needed.

Efficient solution:
Approach: The above solution requires two traversals of linked list. The middle node can delete using one traversal. The idea is to use two pointers, slow_ptr and fast_ptr. Both pointers start from head of list. When fast_ptr reaches 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 previous of middle so the middle node can be deleted.

Below is the implementation.

C++

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// C++ program to delete middle
// of a linked list
#include <bits/stdc++.h>
using namespace std;
  
/* Link list Node */
struct Node {
    int data;
    struct Node* next;
};
  
// Deletes middle node and returns 
// head of the modified list
struct Node* deleteMid(struct Node* head)
{
    // Base cases
    if (head == NULL)
        return NULL;
    if (head->next == NULL) {
        delete head;
        return NULL;
    }
  
    // Initialize slow and fast pointers
    // to reach middle of linked list
    struct Node* slow_ptr = head;
    struct Node* fast_ptr = head;
  
    // Find the middle and previous of middle.
// To store previous of slow_ptr    
struct Node* prev; 
    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;
    delete slow_ptr;
  
    return head;
}
  
// A utility function to print 
// a given linked list
void printList(struct Node* ptr)
{
    while (ptr != NULL) {
        cout << ptr->data << "->";
        ptr = ptr->next;
    }
    cout << "NULL\n";
}
  
// Utility function to create a new node.
Node* newNode(int data)
{
    struct Node* temp = new Node;
    temp->data = data;
    temp->next = NULL;
    return temp;
}
  
/* Driver program to test above function*/
int main()
{
    /* Start with the empty list */
    struct Node* head = newNode(1);
    head->next = newNode(2);
    head->next->next = newNode(3);
    head->next->next->next = newNode(4);
  
    cout << "Gven Linked List\n";
    printList(head);
  
    head = deleteMid(head);
  
    cout << "Linked List after deletion of middle\n";
    printList(head);
  
    return 0;
}

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Java

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// 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;
    }
  
    /* Drier 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("Gven 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.

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C#

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// C# program to delete middle of a linked list
using System;
  
class GfG {
  
    /* Link list Node */
    class Node {
        public int data;
        public 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) {
            Console.Write(ptr.data + "->");
            ptr = ptr.next;
        }
        Console.WriteLine("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;
    }
  
    /* Drier 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);
  
        Console.WriteLine("Gven Linked List");
        printList(head);
  
        head = deleteMid(head);
  
        Console.WriteLine("Linked List after"
                          + "deletion of middle");
        printList(head);
    }
}
  
/* This code is contributed by 29AjayKumar */

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Output:

Gven 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 linked list is needed
  • Auxiliary Space: O(1).
    As no extra space is needed.

This article is contributed by Piyush Gupta. If you like GeeksforGeeks and would like to contribute, you can also write an article and mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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