C 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 of special pointer to remove the required node from the linked list.
C
#include <stdio.h>
#include <stdlib.h>
struct Node
{
int data;
struct Node* next;
};
void deleteLast( struct Node** head,
int x)
{
struct Node** tmp1 = NULL;
while (*head)
{
if ((*head)->data == x)
{
tmp1 = head;
}
head = &(*head)->next;
}
if (tmp1)
{
struct Node* tmp = *tmp1;
*tmp1 = tmp->next;
free (tmp);
}
}
struct Node* newNode( int x)
{
struct Node* node =
malloc ( sizeof ( struct Node*));
node->data = x;
node->next = NULL;
return node;
}
void display( struct Node* head)
{
struct Node* temp = head;
if (head == NULL)
{
printf ( "NULL" );
return ;
}
while (temp != NULL)
{
printf ( "%d --> " ,
temp->data);
temp = temp->next;
}
printf ( "NULL" );
}
int main()
{
struct Node* 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);
printf ( "Created Linked list: " );
display(head);
deleteLast(&head, 4);
printf ( "List after deletion of 4: " );
display(head);
return 0;
}
|
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.
The above solution doesn’t work when the node to be deleted is the last node.
Following solution handles all cases.
C
#include <stdio.h>
#include <stdlib.h>
struct Node
{
int data;
struct Node* next;
};
void deleteLast( struct Node* head,
int x)
{
struct Node *temp = head,
*ptr = NULL;
while (temp)
{
if (temp->data == x)
ptr = temp;
temp = temp->next;
}
if (ptr != NULL &&
ptr->next == NULL)
{
temp = head;
while (temp->next != ptr)
temp = temp->next;
temp->next = NULL;
}
if (ptr != NULL &&
ptr->next != NULL)
{
ptr->data = ptr->next->data;
temp = ptr->next;
ptr->next = ptr->next->next;
free (temp);
}
}
struct Node* newNode( int x)
{
struct Node* node =
malloc ( sizeof ( struct Node*));
node->data = x;
node->next = NULL;
return node;
}
void display( struct Node* head)
{
struct Node* temp = head;
if (head == NULL)
{
printf ( "NULL" );
return ;
}
while (temp != NULL)
{
printf ( "%d --> " , temp->data);
temp = temp->next;
}
printf ( "NULL" );
}
int main()
{
struct Node* 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);
printf ( "Created Linked list: " );
display(head);
deleteLast(head, 4);
printf ( "List after deletion of 4: " );
display(head);
return 0;
}
|
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!
Like Article
Suggest improvement
Share your thoughts in the comments
Please Login to comment...