Coding Linked List PYTHON Singly Linked List

Python Algorithm – Deleting a node in Linked List | Set 3

Python Algorithm - Deleting a node in Linked List - Linked List - We have discussed Linked List Introduction and Linked List Insertion in previous

We have discussed Linked List Introduction and Linked List Insertion in previous posts on singly linked list.

Let us formulate the problem statement to understand the deletion process. Given a ‘key’, delete the first occurrence of this key in linked list.
To delete a node from linked list, we need to do following steps.
1) Find previous node of the node to be deleted.
2) Changed next of previous node.
3) Free memory for the node to be deleted.

Deleting a node in Linked List | Set 3

Since every node of linked list is dynamically allocated using malloc() in C, we need to call free() for freeing memory allocated for the node to be deleted.

Python Programming:

 

# Python program to delete a node from linked list
 
# Node class 
class Node:
 
    # Constructor to initialize the node object
    def __init__(self, data):
        self.data = data
        self.next = None
 
class LinkedList:
 
    # Function to initialize head
    def __init__(self):
        self.head = None
 
    # Function to insert a new node at the beginning
    def push(self, new_data):
        new_node = Node(new_data)
        new_node.next = self.head
        self.head = new_node
 
    # Given a reference to the head of a list and a key,
    # delete the first occurence of key in linked list
    def deleteNode(self, key):
         
        # Store head node
        temp = self.head
 
        # If head node itself holds the key to be deleted
        if (temp is not None):
            if (temp.data == key):
                self.head = temp.next
                temp = None
                return
 
        # Search for the key to be deleted, keep track of the
        # previous node as we need to change 'prev.next'
        while(temp is not None):
            if temp.data == key:
                break
            prev = temp
            temp = temp.next
 
        # if key was not present in linked list
        if(temp == None):
            return
 
        # Unlink the node from linked list
        prev.next = temp.next
 
        temp = None
 
 
    # Utility function to print the linked LinkedList
    def printList(self):
        temp = self.head
        while(temp):
            print " %d" %(temp.data),
            temp = temp.next
 
 
# Driver program
llist = LinkedList()
llist.push(7)
llist.push(1)
llist.push(3)
llist.push(2)
 
print "Created Linked List: "
llist.printList()
llist.deleteNode(1) 
print "\nLinked List after Deletion of 1:"
llist.printList()
 
# This code is contributed by Nikhil Kumar Singh (nickzuck_007)

Output:

Created Linked List:
 2  3  1  7
Linked List after Deletion of 1:
 2  3  7
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About the author

Venkatesan Prabu

Venkatesan Prabu

Wikitechy Founder, Author, International Speaker, and Job Consultant. My role as the CEO of Wikitechy, I help businesses build their next generation digital platforms and help with their product innovation and growth strategy. I'm a frequent speaker at tech conferences and events.

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