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.
<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!