Q. Program to rotate doubly linked list by N nodes.
Explanation
In this program, we need to create a doubly linked list and rotate it by n node. This can be achieved by maintaining a pointer that starts from the head node and traverses the list until current points to the nth node. Move the list from head to the nth node and place it after tail. Now nth node will be the tail of the list and node next to nth node will be the new head. Here, n should always be greater than 0 but less than the size of the list.
Original List:
List after rotating it by 3 nodes:
In the above example, we need to rotate list by 3 nodes. First, we iterate through the list until current points to the 3rd node which is, in this case, are node 3. Move the list from node 1 to 3 and place it after tail. Now, node 4 will be new head and node 3 will be the new tail.
Algorithm
 Define a Node class which represents a node in the list. It will have three properties: data, previous which will point to the previous node and next which will point to the next node.
 Define another class for creating the doubly linked list, and it has two nodes: head and tail. Initially, head and tail will point to null.
 addNode() will add node to the list:
 It first checks whether the head is null, then it will insert the node as the head.
 Both head and tail will point to a newly added node.
 Head's previous pointer will point to null and tail's next pointer will point to null.
 If the head is not null, the new node will be inserted at the end of the list such that new node's previous pointer will point to tail.
 The new node will become the new tail. Tail's next pointer will point to null.
 rotateList() will rotate the list by given n nodes.
 First, check whether n is 0 or greater than or equal to many nodes present in the list.
 If yes, print the list as it is.
 If no, define a node current which will point to head.
 Iterate through the list till current reaches the nth node.
 Tail's next will point to head node.
 Make node next to current as the new head. Head's previous will point to null.
 The current node will become tail of the list. Tail's next will point to null.
 display() will show all the nodes present in the list.
 Define a new node 'current' that will point to the head.
 Print current.data till current points to null.
 Current will point to the next node in the list in each iteration.
Solution
Python
Output:
Original List:
1 2 3 4 5
Updated List:
4 5 1 2 3
C
Output:
Original List:
1 2 3 4 5
Updated List:
4 5 1 2 3
JAVA
Output:
Original List:
1 2 3 4 5
Updated List:
4 5 1 2 3
C#
Output:
Original List:
1 2 3 4 5
Updated List:
4 5 1 2 3
PHP
Output:
Original List:
1 2 3 4 5
Updated List:
4 5 1 2 3
