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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