Python Program For Deleting Last Occurrence Of An Item From Linked List
Last Updated :
22 Jun, 2022
Using pointers, loop through the whole list and keep track of the node prior to the node containing the last occurrence key using a special pointer. After this just store the next of next of the special pointer, into to next special pointer to remove the required node from the linked list.
Python3
# Python program to implement
# the above approach
# A linked list Node
class Node:
def __init__(self, new_data):
self.data = new_data
self.next = None
# Function to delete the last
# occurrence
def deleteLast(head, x):
temp = head
ptr = None
while (temp != None):
# If found key, update
if (temp.data == x):
ptr = temp
temp = temp.next
# If the last occurrence is the
# last node
if (ptr != None and ptr.next == None):
temp = head
while (temp.next != ptr):
temp = temp.next
temp.next = None
# If it is not the last node
if (ptr != None and ptr.next != None):
ptr.data = ptr.next.data
temp = ptr.next
ptr.next = ptr.next.next
return head
# Utility function to create a
# new node
# with given key
def newNode(x):
node = Node(0)
node.data = x
node.next = None
return node
# This function prints contents of
# linked list starting from the given
# Node
def display(head):
temp = head
if (head == None):
print("NULL")
return
while (temp != None):
print(temp.data, " --> ",
end = "")
temp = temp.next
print("NULL")
# Driver code
head = newNode(1)
head.next = newNode(2)
head.next.next = newNode(3)
head.next.next.next =
newNode(4)
head.next.next.next.next =
newNode(5)
head.next.next.next.next.next =
newNode(4)
head.next.next.next.next.next.next =
newNode(4)
print("Created Linked list: ",
end = '')
display(head)
# Pass the address of the head pointer
head = deleteLast(head, 4)
print("List after deletion of 4: ",
end = '')
display(head)
# This code is contributed by rutvik_56
Output:
Created Linked list: 1 --> 2 --> 3 --> 4 --> 5 --> 4 --> 4 --> NULL
List after deletion of 4: 1 --> 2 --> 3 --> 4 --> 5 --> 4 --> NULL
Time Complexity: O(n) where n is the number of nodes in the given linked list.
Auxiliary Space: O(1), no extra space is required, so it is a constant.
Given a linked list and a key to be deleted. Delete last occurrence of key from linked. The list may have duplicates.
Examples:
Input: 1->2->3->5->2->10, key = 2
Output: 1->2->3->5->10
The idea is to traverse the linked list from beginning to end. While traversing, keep track of last occurrence key. After traversing the complete list, delete the last occurrence by copying data of next node and deleting the next node.
Python3
# Python3 program to demonstrate deletion
# of last Node in singly linked list
# A linked list Node
class Node:
# Constructor to initialize the
# node object
def __init__(self, data):
self.data = data
self.next = None
def deleteLast(head, key):
# Initialize previous of Node to
# be deleted
x = None
# Start from head and find the Node
# to be deleted
temp = head
while (temp != None):
# If we found the key, update xv
if (temp.key == key) :
x = temp
temp = temp.next
# key occurs at-least once
if (x != None):
# Copy key of next Node to x
x.key = x.next.key
# Store and unlink next
temp = x.next
x.next = x.next.next
# Free memory for next
return head
# Utility function to create
# a new node with given key
def newNode(key):
temp = Node(0)
temp.key = key
temp.next = None
return temp
# This function prints contents of
# linked list starting from the given
# Node
def printList(node):
while (node != None):
print (node.key,
end = " ")
node = node.next
# Driver Code
if __name__=='__main__':
# Start with the empty list
head = newNode(1)
head.next = newNode(2)
head.next.next = newNode(3)
head.next.next.next =
newNode(5)
head.next.next.next.next =
newNode(2)
head.next.next.next.next.next =
newNode(10)
print("Created Linked List: ")
printList(head)
deleteLast(head, 2)
print("Linked List after Deletion of 1: ")
printList(head)
# This code is contributed by Arnab Kundu
Output:
Created Linked List:
1 2 3 5 2 10
Linked List after Deletion of 1:
1 2 3 5 10
Time Complexity: O(n) where n is the number of nodes in the given linked list.
Auxiliary Space: O(1), no extra space is required, so it is a constant.
The above solution doesn't work when the node to be deleted is the last node.
Following solution handles all cases.
Python3
# A Python3 program to demonstrate deletion
# of last Node in singly linked list
# A linked list Node
class Node:
def __init__(self, new_data):
self.data = new_data
self.next = None
# Function to delete the last
# occurrence
def deleteLast(head, x):
temp = head
ptr = None
while (temp != None):
# If found key, update
if (temp.data == x):
ptr = temp
temp = temp.next
# If the last occurrence is the
# last node
if (ptr != None and ptr.next == None):
temp = head
while (temp.next != ptr) :
temp = temp.next
temp.next = None
# If it is not the last node
if (ptr != None and ptr.next != None):
ptr.data = ptr.next.data
temp = ptr.next
ptr.next = ptr.next.next
return head
# Utility function to create a
# new node with given key
def newNode(x):
node = Node(0)
node.data = x
node.next = None
return node
# This function prints contents of
# linked list starting from the given
# Node
def display(head):
temp = head
if (head == None):
print("None")
return
while (temp != None):
print(temp.data, " -> ",
end = "")
temp = temp.next
print("None")
# Driver code
head = newNode(1)
head.next = newNode(2)
head.next.next = newNode(3)
head.next.next.next =
newNode(4)
head.next.next.next.next =
newNode(5)
head.next.next.next.next.next =
newNode(4)
head.next.next.next.next.next.next =
newNode(4)
print("Created Linked list: ")
display(head)
head = deleteLast(head, 4)
print("List after deletion of 4: ")
display(head)
# This code is contributed by Arnab Kundu
Output:
Created Linked List:
1 2 3 4 5 4 4
Linked List after Deletion of 1:
1 2 3 4 5 4
Time Complexity: O(n) where n is the number of nodes in the given linked list.
Auxiliary Space: O(1), no extra space is required, so it is a constant.
Please refer complete article on Delete last occurrence of an item from linked list for more details!