Python program to remove duplicate elements from a Circular Linked List
In this program, we will create a circular linked list and remove duplicate nodes from the list. We will compare a node with the rest of the list and check for the duplicate. If the duplicate is found, delete the duplicate node from the list.
- 1->2->2->4->3
In the above list, we can see, node 2 is present twice in the list. So, we will have a node current that will iterate through the list. The index will point to the next node to current. Temp will be pointing to the node previous to index. When a duplicate is found, we delete it by pointing temp.next to index.next. Above list after removing duplicates:
- 1->2->4->3
ALGORITHM:
- Define a Node class which represents a node in the list. It has two properties data and next which will point to the next node.
- Define another class for creating the circular linked list, and it has two nodes: head and tail.
- removeDuplicate() will remove duplicate nodes from the list:
- Node current will point to head and used to traverse through the list.
- The index will point to the next node to current and temp will point to the previous node to index.
- We will compare the current.data with the index.data. If the match is found, delete duplicate data by pointing temp’s next to index’s next.
- Increment index to index.next and current to current .next.
- Repeat step from c to d till all the duplicates are removed.
PROGRAM:
- #Represents the node of list.
- #Represents the node of list.
- class Node:
- def __init__(self,data):
- self.data = data;
- self.next = None;
- class CreateList:
- #Declaring head and tail pointer as null.
- def __init__(self):
- self.head = Node(None);
- self.tail = Node(None);
- self.head.next = self.tail;
- self.tail.next = self.head;
- #This function will add the new node at the end of the list.
- def add(self,data):
- newNode = Node(data);
- #Checks if the list is empty.
- if self.head.data is None:
- #If list is empty, both head and tail would point to new node.
- self.head = newNode;
- self.tail = newNode;
- newNode.next = self.head;
- else:
- #tail will point to new node.
- self.tail.next = newNode;
- #New node will become new tail.
- self.tail = newNode;
- #Since, it is circular linked list tail will point to head.
- self.tail.next = self.head;
- #Removes duplicate from the list
- def removeDuplicate(self):
- #Current will point to head
- current = self.head;
- if(self.head == None):
- print(“List is empty”);
- else:
- while(True):
- #Temp will point to previous node of index.
- temp = current;
- #Index will point to node next to current
- index = current.next;
- while(index != self.head):
- #If current node is equal to index data
- if(current.data == index.data):
- #Here, index node is pointing to the node which is duplicate of current node
- #Skips the duplicate node by pointing to next node
- temp.next = index.next;
- else:
- #Temp will point to previous node of index.
- temp = index;
- index= index.next;
- current =current.next;
- if(current.next == self.head):
- break;
- #Displays all the nodes in the list
- def display(self):
- current = self.head;
- if self.head is None:
- print(“List is empty”);
- return;
- else:
- #Prints each node by incrementing pointer.
- print(current.data);
- while(current.next != self.head):
- current = current.next;
- print(current.data);
- print(“\n”);
- class CircularLinkedList:
- cl = CreateList();
- #Adds data to the list
- cl.add(1);
- cl.add(2);
- cl.add(3);
- cl.add(2);
- cl.add(2);
- cl.add(4);
- print(“Originals list: “);
- cl.display();
- #Removes duplicate nodes
- cl.removeDuplicate();
- print(“List after removing duplicates: “);
- cl.display();
Output:
Originals list: 1 2 3 2 2 4 List after removing duplicates: 1 2 3 4