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

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