Delete alternate nodes of a Linked List
Given a Singly Linked List, starting from the second node delete all alternate nodes of it. For example, if the given linked list is 1->2->3->4->5 then your function should convert it to 1->3->5, and if the given linked list is 1->2->3->4 then convert it to 1->3.
Method 1 (Iterative)
Keep track of previous of the node to be deleted. First change the next link of previous node and then free the memory allocated for the node.
C++
// C++ program to remove alternate // nodes of a linked list #include <bits/stdc++.h> using namespace std; /* A linked list node */class Node { public: int data; Node *next; }; /* deletes alternate nodes of a list starting with head */void deleteAlt(Node *head) { if (head == NULL) return; /* Initialize prev and node to be deleted */ Node *prev = head; Node *node = head->next; while (prev != NULL && node != NULL) { /* Change next link of previous node */ prev->next = node->next; /* Free memory */ free(node); /* Update prev and node */ prev = prev->next; if (prev != NULL) node = prev->next; } } /* UTILITY FUNCTIONS TO TEST fun1() and fun2() *//* Given a reference (pointer to pointer) to the head of a list and an int, push a new node on the front of the list. */void push(Node** head_ref, int new_data) { /* allocate node */ Node* new_node = new Node(); /* put in the data */ new_node->data = new_data; /* link the old list off the new node */ new_node->next = (*head_ref); /* move the head to point to the new node */ (*head_ref) = new_node; } /* Function to print nodes in a given linked list */void printList(Node *node) { while (node != NULL) { cout<< node->data<<" "; node = node->next; } } /* Driver code */int main() { /* Start with the empty list */ Node* head = NULL; /* Using push() to construct below list 1->2->3->4->5 */ push(&head, 5); push(&head, 4); push(&head, 3); push(&head, 2); push(&head, 1); cout<<"List before calling deleteAlt() \n"; printList(head); deleteAlt(head); cout<<"\nList after calling deleteAlt() \n"; printList(head); return 0; } // This code is contributed by rathbhupendra |
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C
// C program to remove alternate nodes of a linked list #include<stdio.h> #include<stdlib.h> /* A linked list node */struct Node { int data; struct Node *next; }; /* deletes alternate nodes of a list starting with head */void deleteAlt(struct Node *head) { if (head == NULL) return; /* Initialize prev and node to be deleted */ struct Node *prev = head; struct Node *node = head->next; while (prev != NULL && node != NULL) { /* Change next link of previous node */ prev->next = node->next; /* Free memory */ free(node); /* Update prev and node */ prev = prev->next; if (prev != NULL) node = prev->next; } } /* UTILITY FUNCTIONS TO TEST fun1() and fun2() *//* Given a reference (pointer to pointer) to the head of a list and an int, push a new node on the front of the list. */void push(struct Node** head_ref, int new_data) { /* allocate node */ struct Node* new_node = (struct Node*) malloc(sizeof(struct Node)); /* put in the data */ new_node->data = new_data; /* link the old list off the new node */ new_node->next = (*head_ref); /* move the head to point to the new node */ (*head_ref) = new_node; } /* Function to print nodes in a given linked list */void printList(struct Node *node) { while (node != NULL) { printf("%d ", node->data); node = node->next; } } /* Driver program to test above functions */int main() { /* Start with the empty list */ struct Node* head = NULL; /* Using push() to construct below list 1->2->3->4->5 */ push(&head, 5); push(&head, 4); push(&head, 3); push(&head, 2); push(&head, 1); printf("\nList before calling deleteAlt() \n"); printList(head); deleteAlt(head); printf("\nList after calling deleteAlt() \n"); printList(head); return 0; } |
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Java
// Java program to delete alternate nodes of a linked list class LinkedList { Node head; // head of list /* Linked list Node*/ class Node { int data; Node next; Node(int d) {data = d; next = null; } } void deleteAlt() { if (head == null) return; Node prev = head; Node now = head.next; while (prev != null && now != null) { /* Change next link of previus node */ prev.next = now.next; /* Free node */ now = null; /*Update prev and now */ prev = prev.next; if (prev != null) now = prev.next; } } /* Utility functions */ /* Inserts a new Node at front of the list. */ public void push(int new_data) { /* 1 & 2: Allocate the Node & Put in the data*/ Node new_node = new Node(new_data); /* 3. Make next of new Node as head */ new_node.next = head; /* 4. Move the head to point to new Node */ head = new_node; } /* Function to print linked list */ void printList() { Node temp = head; while(temp != null) { System.out.print(temp.data+" "); temp = temp.next; } System.out.println(); } /* Driver program to test above functions */ public static void main(String args[]) { LinkedList llist = new LinkedList(); /* Constructed Linked List is 1->2->3->4->5->null */ llist.push(5); llist.push(4); llist.push(3); llist.push(2); llist.push(1); System.out.println("Linked List before calling deleteAlt() "); llist.printList(); llist.deleteAlt(); System.out.println("Linked List after calling deleteAlt() "); llist.printList(); } } /* This code is contributed by Rajat Mishra */ |
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Python3
# Python3 program to remove alternate # nodes of a linked list import math # A linked list node class Node: def __init__(self, data): self.data = data self.next = None # deletes alternate nodes # of a list starting with head def deleteAlt(head): if (head == None): return # Initialize prev and node to be deleted prev = head now = head.next while (prev != None and now != None): # Change next link of previous node prev.next = now.next # Free memory now = None # Update prev and node prev = prev.next if (prev != None): now = prev.next # UTILITY FUNCTIONS TO TEST fun1() and fun2() # Given a reference (poer to poer) to the head # of a list and an , push a new node on the front # of the list. def push(head_ref, new_data): # allocate node new_node = Node(new_data) # put in the data new_node.data = new_data # link the old list off the new node new_node.next = head_ref # move the head to po to the new node head_ref = new_node return head_ref # Function to print nodes in a given linked list def prList(node): while (node != None): print(node.data, end = " ") node = node.next # Driver code if __name__=='__main__': # Start with the empty list head = None # Using head=push() to construct below list # 1.2.3.4.5 head = push(head, 5) head = push(head, 4) head = push(head, 3) head = push(head, 2) head = push(head, 1) print("List before calling deleteAlt() ") prList(head) deleteAlt(head) print("\nList after calling deleteAlt() ") prList(head) # This code is contributed by Srathore |
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C#
// C# program to delete alternate // nodes of a linked list using System; public class LinkedList { Node head; // head of list /* Linked list Node*/ public class Node { public int data; public Node next; public Node(int d) { data = d; next = null; } } void deleteAlt() { if (head == null) return; Node prev = head; Node now = head.next; while (prev != null && now != null) { /* Change next link of previus node */ prev.next = now.next; /* Free node */ now = null; /*Update prev and now */ prev = prev.next; if (prev != null) now = prev.next; } } /* Utility functions */ /* Inserts a new Node at front of the list. */ public void push(int new_data) { /* 1 & 2: Allocate the Node & Put in the data*/ Node new_node = new Node(new_data); /* 3. Make next of new Node as head */ new_node.next = head; /* 4. Move the head to point to new Node */ head = new_node; } /* Function to print linked list */ void printList() { Node temp = head; while(temp != null) { Console.Write(temp.data+" "); temp = temp.next; } Console.WriteLine(); } /* Driver code*/ public static void Main(String []args) { LinkedList llist = new LinkedList(); /* Constructed Linked List is 1->2->3->4->5->null */ llist.push(5); llist.push(4); llist.push(3); llist.push(2); llist.push(1); Console.WriteLine("Linked List before" + "calling deleteAlt() "); llist.printList(); llist.deleteAlt(); Console.WriteLine("Linked List after" + "calling deleteAlt() "); llist.printList(); } } // This code has been contributed // by 29AjayKumar |
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Output:
List before calling deleteAlt() 1 2 3 4 5 List after calling deleteAlt() 1 3 5
Time Complexity: O(n) where n is the number of nodes in the given Linked List.
Method 2 (Recursive)
Recursive code uses the same approach as method 1. The recursive coed is simple and short, but causes O(n) recursive function calls for a linked list of size n.
C++
/* deletes alternate nodes of a list starting with head */void deleteAlt(Node *head) { if (head == NULL) return; Node *node = head->next; if (node == NULL) return; /* Change the next link of head */ head->next = node->next; /* free memory allocated for node */ free(node); /* Recursively call for the new next of head */ deleteAlt(head->next); } // This code is contributed by rathbhupendra |
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C
/* deletes alternate nodes of a list starting with head */void deleteAlt(struct Node *head) { if (head == NULL) return; struct Node *node = head->next; if (node == NULL) return; /* Change the next link of head */ head->next = node->next; /* free memory allocated for node */ free(node); /* Recursively call for the new next of head */ deleteAlt(head->next); } |
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Java
/* deletes alternate nodes of a list starting with head */static Node deleteAlt(Node head) { if (head == null) return; Node node = head.next; if (node == null) return; /* Change the next link of head */ head.next = node.next; /* Recursively call for the new next of head */ head.next = deleteAlt(head.next); } // This code is contributed by Arnab Kundu |
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Python3
# deletes alternate nodes of a list starting with head def deleteAlt(head): if (head == None): return node = head.next if (node == None): return # Change the next link of head head.next = node.next # free memory allocated for node #free(node) # Recursively call for the new next of head deleteAlt(head.next) # This code is contributed by Srathore |
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C#
/* deletes alternate nodes of a list starting with head */static Node deleteAlt(Node head) { if (head == null) return; Node node = head.next; if (node == null) return; /* Change the next link of head */ head.next = node.next; /* Recursively call for the new next of head */ head.next = deleteAlt(head.next); } // This code is contributed by Arnab Kundu |
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Time Complexity: O(n)
Please write comments if you find the above code/algorithm incorrect, or find better ways to solve the same problem.
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