# Reverse a doubly linked list

### Problem Statement :

```This challenge is part of a tutorial track by MyCodeSchool

Given the pointer to the head node of a doubly linked list, reverse the order of the nodes in place. That is, change the next and prev pointers of the nodes so that the direction of the list is reversed. Return a reference to the head node of the reversed list.

Note: The head node might be NULL to indicate that the list is empty.

Function Description

Complete the reverse function in the editor below.

reverse has the following parameter(s):

Returns

Input Format

The first line contains an integer t, the number of test cases.

Each test case is of the following format:

The first line contains an integer n, the number of elements in the linked list.
The next n lines contain an integer each denoting an element of the linked list.```

### Solution :

```                            ```Solution in C :

In C :

// Complete the reverse function below.

/*
*
*     int data;
* };
*
*/
prev=NULL;
while(curr)
{
next=curr->next;
curr->next=prev;
curr->prev=next;
if(next==NULL)break;
prev=curr;curr=next;
}
curr->prev=NULL;return curr;

}```
```

```                        ```Solution in C++ :

In C++ :

/*
Reverse a doubly linked list, input list may also be empty
Node is defined as
struct Node
{
int data;
Node *next;
Node *prev
}
*/
{
// Complete this function
// Do not write the main method.

Node *temp = NULL;

/* swap next and prev for all nodes of
while (current !=  NULL)
{
temp = current->prev;
current->prev = current->next;
current->next = temp;
current = current->prev;
}

/* Before changing head, check for the cases like empty
list and list with only one node */
if(temp != NULL )

}```
```

```                        ```Solution in Java :

In Java :

/*
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;
Node prev;
}
*/

return null;

Node next=temp.next;
while(next!=null)
{

temp.next=temp.prev;
temp.prev=next;
temp=next;
next=next.next;

}

temp.next=temp.prev;
temp.prev=null;
return temp;

}```
```

```                        ```Solution in Python :

In python3 :

while curr.next != None:
prev, curr = curr, curr.next

while prev != None:
curr.prev = curr.next
curr.next = prev
prev.next = prev.prev
prev.prev = curr
prev, curr = prev.next, prev

```

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