Python Program For Deleting A Linked List Node At A Given Position

Given a singly linked list and a position, delete a linked list node at the given position.

Example:

Input: position = 1, Linked List = 8->2->3->1->7
Output: Linked List =  8->3->1->7

Input: position = 0, Linked List = 8->2->3->1->7
Output: Linked List = 2->3->1->7

If the node to be deleted is the root, simply delete it. To delete a middle node, we must have a pointer to the node previous to the node to be deleted. So if positions are not zero, we run a loop position-1 times and get a pointer to the previous node.

Below is the implementation of the above idea.

Python

# Python program to delete a node in a linked list
# at a given position

# Node class
class Node:

    # Constructor to initialize the node object
    def __init__(self, data):
        self.data = data
        self.next = None

class LinkedList:

    # Constructor 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 position, delete the node at a given position
    def deleteNode(self, position):

        # If linked list is empty
        if self.head == None:
            return

        # Store head node
        temp = self.head

        # If head needs to be removed
        if position == 0:
            self.head = temp.next
            temp = None
            return

        # Find previous node of the node to be deleted
        for i in range(position -1 ):
            temp = temp.next
            if temp is None:
                break

        # If position is more than number of nodes
        if temp is None:
            return
        if temp.next is None:
            return

        # Node temp.next is the node to be deleted
        # store pointer to the next of node to be deleted
        next = temp.next.next

        # Unlink the node from linked list
        temp.next = None

        temp.next = next


    # Utility function to print the linked LinkedList
    def printList(self):
        temp = self.head
        while(temp):
            print " %d " %(temp.data),
            temp = temp.next


# Driver program to test above function
llist = LinkedList()
llist.push(7)
llist.push(1)
llist.push(3)
llist.push(2)
llist.push(8)

print "Created Linked List: "
llist.printList()
llist.deleteNode(4)
print "
Linked List after Deletion at position 4: "
llist.printList()

# This code is contributed by Nikhil Kumar Singh(nickzuck_007)

Output: 

Created Linked List: 
 8  2  3  1  7 
Linked List after Deletion at position 4: 
 8  2  3  1 

Time Complexity: O(n), where n represents the length of the given linked list.
Auxiliary Space: O(1), no extra space is required, so it is a constant.

Please refer complete article on Delete a Linked List node at a given position for more details!