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->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 fron // 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 fron // 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 fron # 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 fron // 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 |
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 |
Given linked list 85 15 4 20 Reversed Linked list 20 4 15 85
Thanks to Abhay Yadav for suggesting this approach.
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