Python Program For Deleting A Linked List Node At A Given Position
Last Updated :
15 Jun, 2022
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!
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