Q5.

Write a Python program to construct a linked list. Prompt the user for input. Remove any duplicate numbers from the linked list.

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Problem Description

In this program, we need to remove the duplicate nodes from the given singly linked list.

Original List:

ll

List after removing duplicate nodes:

ll

In the above list, node 2 is repeated thrice, and node 1 is repeated twice. Node current will point to head, and index will point to node next to current. Start traversing the list till a duplicate is found that is when current's data is equal to index's data. In the above example, the first duplicate will be found at position 4. Assign current to another node temp. Connect temp's next node with index's next node. Delete index which was pointing to duplicate node. This process will continue until all duplicates are removed.

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Steps for removing duplicate number(s) from the linked list

  1. Create a class Node which has two data attributes: data and next. The data attribute next points to the next node in the list.

  2. Create another class LinkedList which has two data attributes: head and tail.

  3. The add_node() method will add a new node to the list:
    a. Create a new node.
    b. It first checks, whether the head is equal to null which means the list is empty.
    c. If the list is empty, both head and tail will point to a newly added node.
    d. If the list is not empty, the new node will be added to end of the list such that tail's next will point to a newly added node. This new node will become the new tail of the list.

  4. The remove_duplicate() method will remove duplicate nodes from the list.
    a. Define a new node current which will initially point to head.
    b. Node temp will point to current and index will always point to node next to current.
    c. Loop through the list till current points to null.
    d. Check whether current's data is equal to index's data that means index is duplicate of current.
    e. Since index points to duplicate node so skip it by making node next to temp to will point to node next to index, i.e. temp.next = index.next.

  5. The display() method will display the nodes present in the list:
    a. Define a node current which will initially point to the head of the list.
    b. Traverse through the list till current points to null.
    c. Display each node by making current to point to node next to it in each iteration.

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In [1]:
# Represent a node of the singly linked list
class Node:
    def __init__(self, data):
        self.data = data
        self.next = None


class LinkedList:
    # Represent the head and tail of the singly linked list
    def __init__(self):
        self.head = None
        self.tail = None

    # add_node() will add a new node to the list
    def add_node(self, data):
        # Create a new node
        new_node = Node(data)

        # Checks if the list is empty
        if self.head == None:
            # If list is empty, both head and tail will point to new node
            self.head = new_node
            self.tail = new_node
        else:
            # new_node will be added after tail such that tail's next will point to new_node
            self.tail.next = new_node
            # newNode will become new tail of the list
            self.tail = new_node

    # removeDuplicate() will remove duplicate nodes from the list
    def remove_duplicate(self):
        # Node current will point to head
        current = self.head
        index = None
        temp = None

        if self.head == None:
            return
        else:
            while current != None:
                # Node temp will point to previous node to index.
                temp = current
                # Index will point to node next to current
                index = current.next

                while index != None:
                    # If current node's data is equal to index node's 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

    # display() will display all the nodes present in the list
    def display(self):
        # Node current will point to head
        current = self.head
        if self.head == None:
            print("List is empty")
            return
        while current != None:
            # Increment next to print each node
            print(current.data)
            current = current.next


def user_input():
    slist = LinkedList()
    # Add data to the list
    data_list = input("Please enter the elements in the linked list: ").split()
    for data in data_list:
        slist.add_node(int(data))

    # Display original list
    print("Original List: ")
    slist.display()

    # Remove duplicate nodes (if any)
    slist.remove_duplicate()

    print("List after removing duplicate nodes: ")
    slist.display()


if __name__ == "__main__":
    user_input()
Please enter the elements in the linked list: 1 2 3 2 2 4 1
Original List: 
1
2
3
2
2
4
1
List after removing duplicate nodes: 
1
2
3
4