# 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.

## Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

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.

 `// C++ program to delete middle ` `// of a linked list ` `#include ` `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; ` `} `

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++

 `// C++ program to delete middle ` `// of a linked list ` `#include ` `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; ` `} `

## 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; ` `    ``} ` ` `  `    ``/* 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. `

## C#

 `// 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 */`

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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