Remove an element from linked list at a given position - c

Question

I have a simple question about removing an element from a linked list. The only difference between what I am trying to accomplish and what I have seen in code online is that I am trying to remove an element, given a position, rather than given the actual element that needs to be removed.
Any help is appreciated.

Answer 1

You can do that, Here is the sample program which can delete any node based on the index you are supplying as an argumenT.
start -> Pointing to the first node
traverse ->pointing to the node which has to be deleted
traverseNext->pointing to the previous of node which has to be deleted.
And the code looks like below.
#include <iostream>
struct myList
{
int data;
struct myList *next;
};
struct myList *start=NULL;
//this method removes a node from the position index
void remove(int index)
{
myList *traverse = start;
myList *traverseNext = NULL;
int i = 1;
while(i<(index-1))
{
traverse = traverse->next;
i++;
}
traverseNext = traverse;
traverse = traverse->next;
if(traverse->next == NULL)
{
delete traverse;
traverseNext->next = NULL;
traverse = NULL;
return;
}
else
{
traverseNext->next = traverse->next;
delete traverse;
traverse = NULL;
return;
}
}
int main(void)
{
myList *node1;
myList *node2;
myList *node3;
node1 = new myList;
node2 = new myList; //Created 3 nodes of type myList
node3 = new myList;
node1->data = 10;
node1->next = node2;
node2->data = 20;
node2->next = node3;
node3->data = 30;
node3->next = NULL;
start = node1; //start is pointing to node1
remove(2); //removing the node 2, so the output will be 10 30
while(start) //iterating through all the nodes from start, since start
{ //is pointing to the first node.
std::cout<<start->data<<" ";
start = start->next;
}
}

Answer 2

Deleting a node in a linked list to which the pointer is given can be done in O(1) time. We don't have to do traversal.
I am assuming that by position you meant the pointer to the node is given:
Lets say node is the element that needs to be removed.
node->data = node->next->data;
Node* temp = node->next;
node->next = node->next->next;
free(temp);
But if the position means the nth element in the list, the only way would be to traverse up to the (n-1)th element and delete the next element by (regular deletion in a linked list):
Node* temp = previous->next;
previous->next = temp->next;
free(temp);
This is all assuming that the linked-list is a pointer based linked-list

Answer 3

This is simple:
1) Locate the Nth item by iterating through the list, additionally using a counter to keep track of which node you're one.
2) Remove that node, as you would any other linked list.

Answer 4

Seek the list until you find the nth element (use a counter), and then update the previous node's next pointer to point to the one after the one you're currently at (effectively removing it). Adjust the previous pointer if you're using them, too.

Answer 5

If you want to remove multiple items you can first iterate over the list and then collect all items that you want to remove to another list. Then simply call 'removeAll' passing in the collected list.

Answer 6

#include <stdio.h>
#include <stdlib.h>
typedef struct element{
int num;
struct element * next;
} element;
void delNth(element **header, int pos){//pos : zero origin
element *prev, *tmp;
int i;
if(header == NULL || *header == NULL || pos < 0) return ;
if(pos == 0){
tmp = (*header)->next;
free(*header);
*header = tmp;
} else {
prev = *header;
tmp = (*header)->next;
for(i=1;i<pos;++i){
prev = tmp;
tmp = tmp->next;
if(tmp == NULL) return ;//or rise error
}
prev->next = tmp->next;
free(tmp);
}
}
void drop(element *header){
if(header){
drop(header->next);
free(header);
}
}
void printList(element *header){
while (header!=NULL){
printf("%d ",header->num);
header=header->next;
}
printf("\n");
}
int main(int argc, char **argv){
int pos = atoi(argv[1]);
element *a;
element *b;
element *c;
a=malloc(sizeof(element));
b=malloc(sizeof(element));
c=malloc(sizeof(element));
a->num=5;
b->num=6;
c->num=7;
a->next=b;
b->next=c;
c->next=NULL;
printList(a);
delNth(&a, pos);
printList(a);
drop(a);
return 0;
}
/* execute result
>a 0
5 6 7
6 7
>a 1
5 6 7
5 7
>a 2
5 6 7
5 6
>a 3
5 6 7
5 6 7
*/