Reverse a linked list Hackerrank problem solution

You’re given the pointer to the head node of a linked list. Change the next pointers of the nodes so that their order is reversed. The head pointer given may be null meaning that the initial list is empty.

Input Format
You have to complete the Node* Reverse(Node* head) method which takes one argument – the head of the linked list. You should NOT read any input from stdin/console.

Output Format

Change the next pointers of the nodes that their order is reversed and return the head of the reversed linked list. Do NOT print anything to stdout/console.

Sample Input

NULL
2 –> 3 –> NULL

Sample Output

NULL
3 –> 2 –> NULL
Explanation
1. Empty list remains empty
2. List is reversed from 2,3 to 3,2

Solution

To reverse a linked list, iterate through the list and change the next node to the previous node by keeping a track of the previous node for each node.

```/*
The input list will have at least one element
Node is defined as
struct Node
{
int data;
struct Node *next;
}
*/
{
{
}

while(tmp->next!=NULL)
{
tmp=tmp->next;
}
}
```
``````/*
Insert Node at the end of a linked list
head pointer input could be NULL as well for empty list
Node is defined as
class Node {
int data;
Node next;
}
*/
// This is a "method-only" submission.
// You only need to complete this method.

return null;

{
}

Node prev=null;

while(next!=null)
{
curr.next=prev;
prev=curr;
curr=next;
next=next.next;

}

curr.next=prev;
return curr;

}
``````
``````"""
head could be None as well for empty list
Node is defined as

class Node(object):

def __init__(self, data=None, next_node=None):
self.data = data
self.next = next_node

return back the head of the linked list in the below method.
"""

last = None