Open In App

C++ Program To Delete Middle Of Linked List

Last Updated : 23 Jul, 2025
Comments
Improve
Suggest changes
Like Article
Like
Report

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. 

C++14 
// 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";
}

// 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 code
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 << "Given Linked List";
    printList(head);
    head = deleteMid(head);
    cout << "Linked List after deletion of middle";
    printList(head);
    return 0;
}


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.  

C++ 
// 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";
}

// 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 code
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 << "Given Linked List";
    printList(head);
    head = deleteMid(head);
    cout << "Linked List after deletion of middle";
    printList(head);
    return 0;
}

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!

K

kartik
Improve
Article Tags :

Explore

Lightbox