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.
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++14
// C++ program to delete middle// of a linked list#include <bits/stdc++.h>using namespace std;/* Link list Node */struct Node { int data; struct Node* next;};// count of nodesint 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 liststruct 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 listvoid 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;} |
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Python3
# Python3 program to delete middle# of a linked list # Link list Node class Node: def __init__(self): self.data = 0 self.next = None # Count of nodesdef countOfNodes(head): count = 0 while (head != None): head = head.next count += 1 return count# Deletes middle node and returns# head of the modified listdef deleteMid(head): # Base cases if (head == None): return None if (head.next == None): del head return None copyHead = head # Find the count of nodes count = countOfNodes(head) # Find the middle node mid = count // 2 # Delete the middle node while (mid > 1): mid -= 1 head = head.next # Delete the middle node head.next = head.next.next return copyHead# A utility function to print# a given linked listdef printList(ptr): while (ptr != None): print(ptr.data, end = '->') ptr = ptr.next print('NULL') # Utility function to create a new node.def newNode(data): temp = Node() temp.data = data temp.next = None return temp# Driver Codeif __name__=='__main__': # Start with the empty list head = newNode(1) head.next = newNode(2) head.next.next = newNode(3) head.next.next.next = newNode(4) print("Gven Linked List") printList(head) head = deleteMid(head) print("Linked List after deletion of middle") printList(head)# This code is contributed by rutvik_56 |
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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 <bits/stdc++.h>using namespace std;/* Link list Node */struct Node { int data; struct Node* next;};// Deletes middle node and returns // head of the modified liststruct 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 listvoid 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;} |
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Java
// Java program to delete the // middle of a linked listclass 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. |
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Python3
# Python3 program to delete the# middle of a linked list# Linked List Nodeclass Node: def __init__(self, data): self.data = data self.next = None# Create and handle list operationsclass LinkedList: def __init__(self): # Head of the list self.head = None # Add new node to the list end def addToList(self, data): newNode = Node(data) if self.head is None: self.head = newNode return last = self.head while last.next: last = last.next last.next = newNode # Returns the list in string format def __str__(self): linkedListStr = "" temp = self.head while temp: linkedListStr += str(temp.data) + "->" temp = temp.next return linkedListStr + "NULL" # Method deletes middle node def deleteMid(self): # Base cases if (self.head is None or self.head.next is None): return # Initialize slow and fast pointers # to reach middle of linked list slow_Ptr = self.head fast_Ptr = self.head # Find the middle and previous of middle prev = None # To store previous of slow pointer while (fast_Ptr is not None and fast_Ptr.next is not None): fast_Ptr = fast_Ptr.next.next prev = slow_Ptr slow_Ptr = slow_Ptr.next # Delete the middle node prev.next = slow_Ptr.next# Driver codelinkedList = LinkedList()linkedList.addToList(1)linkedList.addToList(2)linkedList.addToList(3)linkedList.addToList(4)print("Given Linked List")print(linkedList)linkedList.deleteMid()print("Linked List after deletion of middle")print(linkedList)# This code is contributed by Debidutta Rath |
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C#
// C# program to delete middle of a linked listusing 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 */ |
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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.
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