Python program to rotate doubly linked list by N nodes.

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Python program to rotate doubly linked list by N nodes.

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 the 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:

Python program to rotate doubly linked list by N nodes

List after rotating it by 3 nodes:

Python program to rotate doubly linked list by N 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:

  1. 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.
  2. 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.
  3. 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.

a. 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 until 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.

a. 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.

PROGRAM:

  1. #Represent a node of doubly linked list  
  2. class Node:
  3.     def __init__(self,data):
  4.         self.data = data;
  5.         self.previous = None;
  6.         self.next = None;
  7. class RotateList:
  8.     #Represent the head and tail of the doubly linked list  
  9.     def __init__(self):
  10.         self.head = None;
  11.         self.tail = None;
  12.         self.size = 0;
  13.     #addNode() will add a node to the list  
  14.     def addNode(self, data):
  15.         #Create a new node  
  16.         newNode = Node(data);
  17.         #If list is empty  
  18.         if(self.head == None):
  19.             #Both head and tail will point to newNode  
  20.             self.head = self.tail = newNode;
  21.             #head’s previous will point to None  
  22.             self.head.previous = None;
  23.             #tail’s next will point to None, as it is the last node of the list  
  24.             self.tail.next = None;
  25.         else:
  26.             #newNode will be added after tail such that tail’s next will point to newNode  
  27.             self.tail.next = newNode;
  28.             #newNode’s previous will point to tail  
  29.             newNode.previous = self.tail;
  30.             #newNode will become new tail  
  31.             self.tail = newNode;
  32.             #As it is last node, tail’s next will point to None  
  33.             self.tail.next = None;
  34.         #Size will count the number of nodes present in the list  
  35.         self.size = self.size + 1;
  36.     #rotateList() will rotate the list by given n nodes  
  37.     def rotateList(self, n):
  38.         #Initially, current will point to head  
  39.         current = self.head;
  40.         #n should not be 0 or greater than or equal to number of nodes present in the list  
  41.         if(n == 0 or n >= self.size):
  42.             return;
  43.         else:
  44.             #Traverse through the list till current point to nth node  
  45.             #after this loop, current will point to nth node  
  46.             for i in range(1, n):
  47.                 current = current.next;
  48.             #Now to move entire list from head to nth node and add it after tail  
  49.             self.tail.next = self.head;
  50.             #Node next to nth node will be new head  
  51.             self.head = current.next;
  52.             #Previous node to head should be None  
  53.             self.head.previous = None;
  54.             #nth node will become new tail of the list  
  55.             self.tail = current;
  56.             #tail’s next will point to None  
  57.             self.tail.next = None;
  58.     #display() will print out the nodes of the list  
  59.     def display(self):
  60.         #Node current will point to head  
  61.         current = self.head;
  62.         if(self.head == None):
  63.             print(“List is empty”);
  64.             return;
  65.         while(current != None):
  66.             #Prints each node by incrementing pointer.  
  67.             print(current.data),
  68.             current = current.next;
  69.         print();
  70. dList = RotateList();
  71. #Add nodes to the list  
  72. dList.addNode(1);
  73. dList.addNode(2);
  74. dList.addNode(3);
  75. dList.addNode(4);
  76. dList.addNode(5);
  77. print(“Original List: “);
  78. dList.display();
  79. #Rotates list by 3 nodes  
  80. dList.rotateList(3);
  81. print(“Updated List: “);
  82. dList.display();

Output:

Original List: 
1 2 3 4 5 
Updated List: 
4 5 1 2 3

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