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Iteratively Reverse a linked list using only 2 pointers (An Interesting Method)

  • Difficulty Level : Medium
  • Last Updated : 26 Oct, 2021

Given pointer to the head node of a linked list, the task is to reverse the linked list.

Examples: 

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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->NULL

We 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 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 */
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 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 rutvik_56

Python




# Iteratively Reverse a linked list using only 2 pointers (An Interesting Method)
# Python program to reverse a linked list
# Link list node 
# node class
class node:
  
    # Constructor to initialize the node object
    def __init__(self, data):
        self.data = data
        self.next = None
  
class 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 off 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,
            temp = temp.next
  
  
# Driver program to test above functions
llist = 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 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 pratham76

Javascript




<script>
// javascript program to reverse a linked
// list using two pointers.
 
    // Link list node
class 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 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 todaysgaurav.
</script>
Output: 
Given linked list
85  15  4  20  
Reversed Linked list 
20  4  15  85

 

Time Complexity: O(n) 
Reference : 
http://discuss.joelonsoftware.com/default.asp?interview.11.564944.16 
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;
}

Python3




# Python3 program to reverse a linked list using two pointers.
 
# A linked list node
class Node :
    def __init__(self):
        self.data = 0
        self.next = None
 
# Function to reverse the linked list using 2 pointers
def 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 node
def 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 list
def printList( head):
 
    temp = head
    while (temp != None) :
        print( temp.data, end=" ")
        temp = temp.next
     
# Driver code
 
# Start with the empty list
head = None
 
head = 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
Output: 
Given linked list
85  15  4  20  
Reversed Linked list 
20  4  15  85

 

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.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
 




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