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 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 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.