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.
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.
Javascript
<script> // Javascript program to delete the // middle of a linked list // Link list Node class Node { constructor() { this.data = 0; this.next = null; } } // Deletes middle node and returns // head of the modified list function deleteMid( 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 var slow_ptr = head; var fast_ptr = head; // Find the middle and previous of // middle. var 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 function printList(ptr) { while (ptr != null) { document.write(ptr.data + "->"); ptr = ptr.next; } document.write("NULL<br/>"); } // Utility function to create a // new node. function newNode(data) { temp = new Node(); temp.data = data; temp.next = null; return temp; } // Driver code // Start with the empty list head = newNode(1); head.next = newNode(2); head.next.next = newNode(3); head.next.next.next = newNode(4); document.write("Given Linked List<br/>"); printList(head); head = deleteMid(head); document.write( "Linked List after deletion of middle<br/>"); printList(head); // This code is contributed by umadevi9616 </script> |
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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