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

3.4

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++

// 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 http://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)
Space Complexity: O(1)

Reference :
http://discuss.joelonsoftware.com/default.asp?interview.11.564944.16

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.

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