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->NULL

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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++

// 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; 
} 

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 

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++ 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; 
} 

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 contribute.geeksforgeeks.org or mail your article to contribute@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.

My Personal Notes

Suggest a Topic

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C/C++

Python

Recommended Posts: