Finding and removing the last occurrence of an element in a (singly) linked list with only one traversal - data-structures

Question

Is it possible to find the last occurrence of an element (for example, an integer) and remove this node with only one (forward) traversal through the list?

Answer 1

Yes.
Simply remember the previous entry every time you find the value you're searching for on the traversal. When the traversal is complete, the last entry remembered will have a link to the entry to be removed, and that is sufficient to do the removal.

Answer 2

public void DeleteLastOccurenceOfKey(Node head, int key)
{
Node current=head;
Node prev=null;
Node temp=null;
while(current!=null)
{
if(current.next!=null && current.next.data==key)
{
prev=current;
temp=current.next;
}
current=current.next;
}
prev.next=temp.next;
}
DeleteLastOccurenceOfKey(head,25);
I/P:5 10 15 25 35 25 40
O/P:5 10 15 25 35 40

Answer 3

/*
* Delete last occurrence of an item from linked list
* Given a liked 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`enter code here`
* Output: 1->2->3->5->10
*/
#include <stdio.h>
#include <stdlib.h>
#include <assert.h>
typedef struct list_ list;
struct list_ {
int d;
list *next;
};
void insert (list **head, int d) {
list *tmp = (list *)malloc(sizeof(list));
assert(tmp);
tmp->d = d;
tmp->next = *head;
*head = tmp;
}
void printL (list *p) {
while (p) {
printf (" %d ", p->d);
p = p->next;
}
printf ("\n");
}
void deletlastOccurence (list **head, int d) {
list *cur = *head;
list *prev = NULL;
list *match = NULL;
if (cur == NULL) {
printf ("list is empty\n");
return;
}
/*
* Special case when there only ONE NODE
* in the LIST
*/
if (cur->next == NULL) {
if (cur->d == d) {
printf ("Deleted one node %d\n", cur->d);
free(cur);
*head = NULL;
} else {
printf(" No match\n");
}
return;
}
/*
* Keep track of previous node
*/
while (cur && cur->next) {
if (cur->next->d == d) {
prev = cur;
match = cur->next;
}
cur = cur->next;
}
if (prev){
prev->next = match->next;
printf ("Delete %d\n", match->d);
free (match);
} else {
/*
* Special case when the last node is
* on the head itself
*/
if ((*head)->d == d) {
cur = *head;
*head = cur->next;
printf("element is at head Delete %d\n", cur->d);
free (cur);
} else {
printf ("No match\n");
}
}
printL(*head);
}
int main (int argc , char *argv) {
list *h = NULL;
insert(&h, 1);
insert(&h, 2);
insert(&h, 3);
insert(&h, 4);
insert(&h, 5);
insert(&h, 2);
insert(&h, 1);
insert(&h, 6);
printL(h);
deletlastOccurence(&h, 6);
deletlastOccurence(&h, 2);
}

Answer 4

public void deleteLastOccurence(int value) {
Element cur = this.head;
Element prev = null;
Element tmp = null;
if(this.head == null)
return;
if(this.head.data == value) {
this.head = null;
return;
}
while(cur != null) {
if(cur.next != null && cur.next.data == value) {
prev = cur;
tmp = cur.next;
}
cur = cur.next;
}
prev.next = tmp.next;
}