Iteratively Reverse a linked list using only 2 pointers (An Interesting Method)
Given pointer to the head node of a linked list, the task is to reverse the linked list.
Examples:
Input : Head of following linked list
1->2->3->4->NULL
Output : Linked list should be changed to,
4->3->2->1->NULL
Input : Head of following linked list
1->2->3->4->5->NULL
Output : Linked list should be changed to,
5->4->3->2->1->NULLWe have seen how to reverse a linked list in article Reverse a linked list. In iterative method we had used 3 pointers prev, cur and next. Below is an interesting approach that uses only two pointers. The idea is to use XOR to swap pointers.
C++
// C++ program to reverse a linked list using two pointers.#include <bits/stdc++.h>using namespace std;typedef uintptr_t ut;/* Link list node */struct Node { int data; struct Node* next;};/* Function to reverse the linked list using 2 pointers */void reverse(struct Node** head_ref){ struct Node* prev = NULL; struct Node* current = *head_ref; // at last prev points to new head while (current != NULL) { // This expression evaluates from left to right // current->next = prev, changes the link from // next to prev node // prev = current, moves prev to current node for // next reversal of node // This example of list will clear it more // 1->2->3->4 initially prev = 1, current = 2 Final // expression will be current = 1^2^3^2^1, as we // know that bitwise XOR of two same numbers will // always be 0 i.e; 1^1 = 2^2 = 0 After the // evaluation of expression current = 3 that means // it has been moved by one node from its previous // position current = (struct Node*)((ut)prev ^ (ut)current ^ (ut)(current->next) ^ (ut)(current->next = prev) ^ (ut)(prev = current)); } *head_ref = prev;}/* Function to push a node */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 of the new node */ new_node->next = (*head_ref); /* move the head to point to the new node */ (*head_ref) = new_node;}/* Function to print linked list */void printList(struct Node* head){ struct Node* temp = head; while (temp != NULL) { printf("%d ", temp->data); temp = temp->next; }}/* Driver program to test above function*/int main(){ /* Start with the empty list */ struct Node* head = NULL; push(&head, 20); push(&head, 4); push(&head, 15); push(&head, 85); printf("Given linked list\n"); printList(head); reverse(&head); printf("\nReversed Linked list \n"); printList(head); return 0;} |
Java
// Java program to reverse a linked// list using two pointers.import java.util.*;class Main { // Link list node static class Node { int data; Node next; }; static Node head_ref = null; // Function to reverse the linked // list using 2 pointers static void reverse() { Node prev = null; Node current = head_ref; // At last prev points to new head while (current != null) { // This expression evaluates from left to right // current.next = prev, changes the link from // next to prev node // prev = current, moves prev to current node // for next reversal of node This example of // list will clear it more 1.2.3.4 initially // prev = 1, current = 2 Final expression will // be current = 1^2^3^2^1, as we know that // bitwise XOR of two same numbers will always // be 0 i.e; 1^1 = 2^2 = 0 After the evaluation // of expression current = 3 that means it has // been moved by one node from its previous // position Node next = current.next; current.next = prev; prev = current; current = next; } head_ref = prev; } // Function to push a node static void push(int new_data) { // Allocate node Node new_node = new Node(); // Put in the data new_node.data = new_data; // Link the old list of the new node new_node.next = (head_ref); // Move the head to point to the new node (head_ref) = new_node; } // Function to print linked list static void printList() { Node temp = head_ref; while (temp != null) { System.out.print(temp.data + " "); temp = temp.next; } } // Driver code public static void main(String[] args) { push(20); push(4); push(15); push(85); System.out.print("Given linked list\n"); printList(); reverse(); System.out.print("\nReversed Linked list \n"); printList(); }}// This code is contributed by rutvik_56 |
Python3
# Iteratively Reverse a linked list using only 2 pointers (An Interesting Method)# Python program to reverse a linked list# Link list node# node classclass node: # Constructor to initialize the node object def __init__(self, data): self.data = data self.next = Noneclass LinkedList: # Function to initialize head def __init__(self): self.head = None # Function to reverse the linked list def reverse(self): prev = None current = self.head # Described here https://www.geeksforgeeks.org/ # how-to-swap-two-variables-in-one-line / while(current is not None): # This expression evaluates from left to right # current->next = prev, changes the link from # next to prev node # prev = current, moves prev to current node for # next reversal of node # This example of list will clear it more 1->2 # initially prev = 1, current = 2 # Final expression will be current = 1, prev = 2 next, current.next = current.next, prev prev, current = current, next self.head = prev # Function to push a new node def push(self, new_data): # allocate node and put in the data new_node = node(new_data) # link the old list of the new node new_node.next = self.head # move the head to point to the new node self.head = new_node # Function to print the linked list def printList(self): temp = self.head while(temp): print(temp.data, end=" ") temp = temp.next# Driver program to test above functionsllist = LinkedList()llist.push(20)llist.push(4)llist.push(15)llist.push(85)print("Given Linked List")llist.printList()llist.reverse()print("\nReversed Linked List")llist.printList()# This code is contributed by Afzal Ansari |
C#
// C# program to reverse a linked// list using two pointers.using System;using System.Collections;using System.Collections.Generic;class GFG { // Link list node class Node { public int data; public Node next; }; static Node head_ref = null; // Function to reverse the linked // list using 2 pointers static void reverse() { Node prev = null; Node current = head_ref; // At last prev points to new head while (current != null) { // This expression evaluates from left to right // current.next = prev, changes the link from // next to prev node // prev = current, moves prev to current node // for next reversal of node This example of // list will clear it more 1.2.3.4 initially // prev = 1, current = 2 Final expression will // be current = 1^2^3^2^1, as we know that // bitwise XOR of two same numbers will always // be 0 i.e; 1^1 = 2^2 = 0 After the evaluation // of expression current = 3 that means it has // been moved by one node from its previous // position Node next = current.next; current.next = prev; prev = current; current = next; } head_ref = prev; } // Function to push a node static void push(int new_data) { // Allocate node Node new_node = new Node(); // Put in the data new_node.data = new_data; // Link the old list of the new node new_node.next = (head_ref); // Move the head to point to the new node (head_ref) = new_node; } // Function to print linked list static void printList() { Node temp = head_ref; while (temp != null) { Console.Write(temp.data + " "); temp = temp.next; } } // Driver code public static void Main(string[] args) { push(20); push(4); push(15); push(85); Console.Write("Given linked list\n"); printList(); reverse(); Console.Write("\nReversed Linked list \n"); printList(); }}// This code is contributed by pratham76 |
Javascript
<script>// javascript program to reverse a linked// list using two pointers. // Link list nodeclass Node{ constructor() { this.data = 0; this.next = null; }} var head_ref = null; // Function to reverse the linked // list using 2 pointers function reverse() { var prev = null; var current = head_ref; // At last prev points to new head while (current != null) { // This expression evaluates from left to right // current.next = prev, changes the link from // next to prev node // prev = current, moves prev to current node for // next reversal of node // This example of list will clear it more 1.2.3.4 // initially prev = 1, current = 2 // Final expression will be current = 1^2^3^2^1, // as we know that bitwise XOR of two same // numbers will always be 0 i.e; 1^1 = 2^2 = 0 // After the evaluation of expression current = 3 that // means it has been moved by one node from its // previous position var next = current.next; current.next = prev; prev = current; current = next; } head_ref = prev; } // Function to push a node function push(new_data) { // Allocate node var new_node = new Node(); // Put in the data new_node.data = new_data; // Link the old list of the new node new_node.next = (head_ref); // Move the head to point to the new node (head_ref) = new_node; } // Function to print linked list function printList() { var temp = head_ref; while (temp != null) { document.write(temp.data + " "); temp = temp.next; } } // Driver code push(20); push(4); push(15); push(85); document.write("Given linked list<br/>"); printList(); reverse(); document.write("<br/>Reversed Linked list <br/>"); printList();// This code is contributed by todaysgaurav.</script> |
Output:
Given linked list 85 15 4 20 Reversed Linked list 20 4 15 85
Time Complexity: O(n)
Auxiliary Space: O(1) since using space for prev and next
Alternate Solution :
C++
// C++ program to reverse a linked list using two pointers.#include <bits/stdc++.h>using namespace std;typedef uintptr_t ut;/* Link list node */struct Node { int data; struct Node* next;};/* Function to reverse the linked list using 2 pointers */void reverse(struct Node** head_ref){ struct Node* current = *head_ref; struct Node* next; while (current->next != NULL) { next = current->next; current->next = next->next; next->next = (*head_ref); *head_ref = next; }}/* Function to push a node */void push(struct Node** head_ref, int new_data){ struct Node* new_node = new Node; new_node->data = new_data; new_node->next = (*head_ref); (*head_ref) = new_node;}/* Function to print linked list */void printList(struct Node* head){ struct Node* temp = head; while (temp != NULL) { printf("%d ", temp->data); temp = temp->next; }}/* Driver program to test above function*/int main(){ /* Start with the empty list */ struct Node* head = NULL; push(&head, 20); push(&head, 4); push(&head, 15); push(&head, 85); printf("Given linked list\n"); printList(head); reverse(&head); printf("\nReversed Linked list \n"); printList(head); return 0;} |
Java
// Java program to reverse a linked// list using two pointers.import java.util.*;class Main { // Link list node static class Node { int data; Node next; }; static Node head_ref = null; // Function to reverse the linked // list using 2 pointers static void reverse() { Node next; Node curr = head_ref; while (curr.next != null) { next = curr.next; curr.next = next.next; next.next = head_ref; head_ref = next; } } // Function to push a node static void push(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 linked list static void printList() { Node temp = head_ref; while (temp != null) { System.out.print(temp.data + " "); temp = temp.next; } } // Driver code public static void main(String[] args) { push(20); push(4); push(15); push(85); System.out.print("Given linked list\n"); printList(); reverse(); System.out.print("\nReversed Linked list \n"); printList(); }}// This code is contributed by Abhijeet Kumar(abhijeet19403) |
Python3
# Python3 program to reverse a linked list using two pointers.# A linked list nodeclass Node : def __init__(self): self.data = 0 self.next = None# Function to reverse the linked list using 2 pointersdef reverse(head_ref): current = head_ref next= None while (current.next != None) : next = current.next current.next = next.next next.next = (head_ref) head_ref = next return head_ref# Function to push a nodedef push( head_ref, new_data): new_node = Node() new_node.data = new_data new_node.next = (head_ref) (head_ref) = new_node return head_ref# Function to print linked listdef printList( head): temp = head while (temp != None) : print( temp.data, end=" ") temp = temp.next # Driver code# Start with the empty listhead = Nonehead = push(head, 20)head = push(head, 4)head = push(head, 15)head = push(head, 85)print("Given linked list")printList(head)head = reverse(head)print("\nReversed Linked list ")printList(head)# This code is contributed by Arnab Kundu |
C#
// C# program to reverse a linked// list using two pointers.using System;using System.Collections;using System.Collections.Generic;class GFG { // Link list node class Node { public int data; public Node next; }; static Node head_ref = null; // Function to reverse the linked // list using 2 pointers static void reverse() { Node current = head_ref; Node next; while (current.next != null) { next = current.next; current.next = next.next; next.next = head_ref; head_ref = next; } } // Function to push a node static void push(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 linked list static void printList() { Node temp = head_ref; while (temp != null) { Console.Write(temp.data + " "); temp = temp.next; } } // Driver code public static void Main(string[] args) { push(20); push(4); push(15); push(85); Console.Write("Given linked list\n"); printList(); reverse(); Console.Write("\nReversed Linked list \n"); printList(); }}// This code is contributed by Abhijeet Kumar(abhijeet19403) |
Javascript
<script>// javascript program to reverse a linked// list using two pointers. // Link list nodeclass Node{ constructor() { this.data = 0; this.next = null; }} var head_ref = null; // Function to reverse the linked // list using 2 pointers function reverse() { var current = head_ref; var next; while (current.next != null) { next = current.next; current.next = next.next; next.next = head_ref; head_ref = next; } } // Function to push a node function push(new_data) { // Allocate node var 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 linked list function printList() { var temp = head_ref; while (temp != null) { document.write(temp.data + " "); temp = temp.next; } } // Driver code push(20); push(4); push(15); push(85); document.write("Given linked list<br/>"); printList(); reverse(); document.write("<br/>Reversed Linked list <br/>"); printList();// This code is contributed by Abhijeet Kumar(abhijeet19403)</script> |
Output:
Given linked list 85 15 4 20 Reversed Linked list 20 4 15 85
Time Complexity : O(n)
Auxiliary Space: O(1)
Thanks to Abhay Yadav for suggesting this approach.
This article is contributed by Shashank Mishra ( Gullu ). If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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