Linked List Delete Function Modification Issues in c - c

I am studying for a test and I am having trouble debugging my linked list practice exam questions. I am having issues with the different modification to the delete functions he wants us to do. Specifically, my deleteEven is an endless look and my deleteNthNode does not work as it is supposed to. Because I am struggling with these two I'm not even sure where to start with deleting every other element in the list. I'd appreciate the help as I want to completely understand this topic before I move to stacks and queues! :)
Below is my code:
/*
Q1:
Try implementing the functions shown in class on your own:
check: node creation
check: insertion at the end of a linked list,
check: insertion at the head of a linked list,
check: a list printing function.
Q2:
check: Write a recursive printList() function.
Q3:
check: Write a recursive tailInsert() function.
Q4:
check: Write a function that inserts nodes at the beginning of the linked list.
Q5:
check: Write a recursive function that prints a linked list in reverse order.
The function signature is: void printReverse(node *head);
Q6:
check: Write an iterative destroyList() function that frees all the nodes in a linked list.
Q7:
check: Now implement destroyList() recursively.
Q8:
- Write a function that deletes the nth element from a linked list.
If the linked list doesn't even have n nodes, don't delete any of them.
The function signature is: node *deleteNth(node *head, int n).
- Try implementing the function iteratively and recursively.
- (In terms of how to interpret n, you can start counting your nodes from zero or one; your choice.)
Q9:
- Write a function that deletes every other element in a linked list.
- (Try writing it both ways: one where it starts deleting at the head of the list,
- and one where it starts deleting at the element just after the head of the list.)
- Can you write this both iteratively and recursively?
Q10:
- Write a function that deletes all even integers from a linked list.
Q11:
- Write a function that takes a sorted linked list and an element to be inserted into that linked list,
and inserts the element in sorted order.
The function signature is: node *insertSorted(node *head, int n);
Q12:
- One of the problems with the first insertNode() function from today is that it
requires us to call it using head = insertNode(head, i).
That's a bit dangerous, because we could forget the "head =" part very easily.
Re-write the function so that it takes a pointer to head,
thereby allowing it to directly modify the contents of head without any need for a return value.
The function signature is: void insertNode(node **head, int data).
The function will be called using insertNode(&head, i).
*/
//come back to
#include <stdio.h>
#include <stdlib.h>
// Basic linked list node struct; contains 'data' and 'next' pointer.
// What happens if we type "node *next" instead of "struct node *next"?
typedef struct node
{
// data field
int data;
// the next node in the list
struct node *next;
} node;
// Allocate a new node. Initialize its fields. Return the pointer.
// We call this from our insertion functions.
node *createNode(int data)
{
node *ptr = NULL;
ptr = malloc(sizeof(node));
if(ptr == NULL)
{
printf("space could not be allocated\n");
return NULL;
}
ptr->data = data;
ptr->next = NULL;
return ptr;
}
// Insert into the end of the linked list. Return the head of the linked list.
// (What is the order (Big-Oh) of this function?)
/*
node *insertNode(node *head, int data)
{
node *temp;
if (head == NULL)
return createNode(data);
for(temp = head; temp->next != NULL; temp = temp->next)
;
temp->next = createNode(data);
return head;
}
*/
node *insertNodeFront(node *head, int data)
{
node *temp;
if(head == NULL)
return createNode(data);
temp = createNode(data);
temp->next = head;
return temp;
}
// Simple function to print the contents of a linked list.
void printList(node *head)
{
if (head == NULL)
{
printf("Empty List\n");
return;
}
for(; head != NULL; head = head->next)
printf("%d ", head->data);
printf("\n");
}
void printListRecursiveHelper(node *head)
{
if (head == NULL)
return;
printf("%d%c", head->data, (head->next == NULL) ? '\n' : ' ');
printListRecursiveHelper(head->next);
}
void printListRecursive(node *head)
{
if (head == NULL)
{
printf("empty list\n");
return;
}
printListRecursiveHelper(head);
}
// Q3: - Write a recursive tailInsert() function.
node *tailInsert(node *head, int data)
{
if(head->next == NULL)
{
node *temp;
temp = createNode(data);
temp->next = NULL;
head->next = temp;
return temp;
}
return tailInsert(head->next, data);
}
//Q5: Write a recursive function that prints a linked list in reverse order.
void printReverse(node *head)
{
if (head == NULL)
return;
printReverse(head->next);
printf("%d ", head->data);
}
// Q6: - Write an iterative destroyList() function that frees all the nodes in a linked list.
// Got code from internet, memorize it
/* Function to delete the entire linked list */
void destroyList (struct node** head)
{
struct node* current = *head;
struct node* next;
while (current != NULL)
{
next = current->next;
free(current);
current = next;
}
*head = NULL;
}
// Q7: - Now implement destroyList() recursively.
// Look up online, need to examine why it deson't work
node *destroyListRecursive(node * head)
{
if (head != NULL)
{
destroyListRecursive(head->next);
free(head);
}
return NULL;
}
/* Q8:
- Write a function that deletes the nth element from a linked list.
- If the linked list doesn't even have n nodes, don't delete any of them.
- Try implementing the function iteratively and recursively.
- (In terms of how to interpret n, you can start counting your nodes from zero or one; your choice.)
*/
node *deleteNth(node *head, int n)
{
/*
int i;
node*current;
node *prev;
current = head;
while(i = 0; i < n; i++)
{
current = current->next;
}
prev = current;
current = head->next->next;
*/
return head;
}
/*
Q9:
- Write a function that deletes every other element in a linked list.
- (Try writing it both ways: one where it starts deleting at the head of the list,
- and one where it starts deleting at the element just after the head of the list.)
- Can you write this both iteratively and recursively?
*/
/* deletes alternate nodes of a list starting with head */
//got code from http://www.geeksforgeeks.org/delete-alternate-nodes-of-a-linked-list/
void deleteAltHead(struct node *head)
{
if (head == NULL)
return;
struct node *node = head->next;
if (node == NULL)
return;
/* Change the next link of head */
head->next = node->next;
/* free memory allocated for node */
free(node);
/* Recursively call for the new next of head */
deleteAltHead(head->next);
}
void deleteOtherTail()
{
}
//Q10: - Write a function that deletes all even integers from a linked list.
void deleteEvenInts(node **head)
{
}
// Q11: - Write a function that takes a sorted linked list and an element to be inserted into that linked list,
// and inserts the element in sorted order.
node *insertSorted(node *head, int n)
{
struct node *current;
struct node *newNode = createNode(n);
/* Special case for the head end */
if (head == NULL || head->data >= newNode->data)
{
newNode->next = head;
head = newNode;
}
else
{
/* Locate the node before the point of insertion */
current = head;
while (current->next != NULL &&
current->next->data < newNode->data)
{
current = current->next;
}
newNode->next = current->next;
current->next = newNode;
}
return current;
//or should it return head?
}
// Q12: Re-write the insertNode() function so that it takes a pointer to head
void insertNode(node **head, int data)
{
node *newNode = createNode(data);
while (*head != NULL)
{
head = &(*head)->next;
}
newNode->next = *head;
*head = newNode;
}
// Q10: Write a function that deletes all even integers from a linked list.
void deleteEven(node **head)
{
node *current = *head;
node *next = current;
node *prev = NULL;
for ( ; current != NULL; )
{
next = current->next;
if( current->data %2)
{
if(prev != NULL)
{
prev->next = next;
free(current);
current = next;
}
else
{
free(current);
current = next;
}
}
else
{
prev = current;
current = next;
}
}
}
//Q8: Write a function that deletes the nth element from a linked list. If the linked list doesn't even have n nodes, don't delete any of them.
node *deleteNthNode(node *head, int n)
{
int i = 0;
int nBigger = 0;
node *current = head;
node *prev = current;
for ( i = 0; i < n; i++)
{
prev = current;
current = current->next;
if (current == NULL)
{
nBigger = 1;
}
}
if(nBigger == 1)
{
return head;
}
prev = current->next;
free(current);
return head;
}
int main(void)
{
int i, r;
// The head of our linked list. If we don't initialize it to NULL, our
// insertNode() function might segfault.
node *head = NULL;
srand(time(NULL));
// Populate the linked list with random integers. We are inserting into the
// head of the list each time.
for (i = 0; i < 10; i++)
{
printf("Inserting %d...\n", r = rand() % 20 + 1);
insertNode(&head, r);
}
head = insertNodeFront(head, 1);
tailInsert(head, 5);
// Print the linked list.
printList(head);
printf("\n");
printReverse(head);
printf("\n\n");
// Print the linked list using our recursive function.
printListRecursive(head);
//destroyList(&head);
deleteAltHead(head);
printf("\n");
printList(head);
{
if (head == NULL)
return;
struct node *node = head->next;
if (node == NULL)
return;
/* Change the next link of head */
head->next = node->next;
/* free memory allocated for node */
free(node);
/* Recursively call for the new next of head */
}
deleteEven(&head);
// head = destroyListRecursive(head);
printf("\n");
printList(head);
node *head2 = NULL;
insertNode(&head2, 1);
insertNode(&head2, 2);
insertNode(&head2, 4);
printf("\n");
printList(head2);
insertSorted(head2, 3);
printf("\n");
deleteNthNode(head, 1);
printList(head2);
//destroyListRecursive(head2);
system("PAUSE");
return 0;
}

Related

Removing unique elements in a doubly linked list in C

I need a little help removing unique characters in a doubly linked list in C. So here's the logic I tried implementing: I counted the occurrence of each character in the doubly linked list. If it's occurrence is 1 time, then it is unique element and needs to be deleted. I'll be repeating the process for all elements. But my code in remove_unique_dll() function isn't working properly, please help me fix it. Here's my code-
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
struct node
{
char data;
struct node *next;
struct node *prev;
};
struct node *head, *tail = NULL; //Represent the head and tail of the doubly linked list
int len;
void addNode(char data)
{
struct node *newNode = (struct node*) malloc(sizeof(struct node)); //Create new node
newNode->data = data;
if (head == NULL)
{ //If dll is empty
head = tail = newNode; //Both head and tail will point to newNode
head->prev = NULL; //head's previous will point to NULL
tail->next = NULL; //tail's next will point to NULL, as it is the last node of the list
}
else
{
tail->next = newNode; //newNode will be added after tail such that tail's next points to newNode
newNode->prev = tail; //newNode's previous will point to tail
tail = newNode; //newNode will become new tail
tail->next = NULL; //As it is last node, tail's next will point to NULL
}
}
void remove_unique_dll()
{
struct node *current = head;
struct node *next;
struct node *prev;
int cnt;
while (current != NULL)
{
next = current->next;
cnt = 1;
//printf("!%c ",next->data);
while (next != NULL)
{
if (next->data == current->data)
{
cnt += 1;
next = next->next;
}
else
next = next->next;
//printf("#%c %d %c\n",next->data,cnt,current->data);
}
if (cnt == 1)
{
prev = current->prev;
//printf("#%c %d",prev->data,cnt);
if (prev == NULL)
{
head = next;
}
else
{
prev->next = next;
}
if (next == NULL)
{
tail = prev;
}
else
{
next->prev = prev;
}
}
current = current->next;
//printf("#%c ",current->data);
}
head = current;
}
void display()
{
struct node *current = head; //head the global one
while (current != NULL)
{
printf("%c<->", current->data); //Prints each node by incrementing pointer.
current = current->next;
}
printf("NULL\n");
}
int main()
{
char s[100];
int i;
printf("Enter string: ");
scanf("%s", s);
len = strlen(s);
for (i = 0; i < len; i++)
{
addNode(s[i]);
}
printf("Doubly linked list: \n");
display();
remove_unique_dll();
printf("Doubly linked list after removing unique elements: \n");
display();
return 0;
}
The output is like this-
If you uncomment the printf() statements inside remove_unique_dll() you'll notice that no code below inner while loop is being executed after inner while loop ends. What's the issue here and what's the solution?
Sample input- aacb
Expected output- a<->a<->NULL
Some issues:
You shouldn't assign head = current at the end, because by then current is NULL
The next you use in the deletion part is not the successor of current, so this will make wrong links
As you progress through the list, every value is going to be regarded as unique at some point: when it is the last occurrence, you'll not find a duplicate anymore, as your logic only looks ahead, not backwards.
When you remove a node, you should free its memory.
Not a big issue, but there is no reason to really count the number of duplicates. Once you find the first duplicate, there is no reason to look for another.
You should really isolate the different steps of the algorithm in separate functions, so you can debug and test each of those features separately and also better understand your code.
Also, to check for duplicates, you might want to use the following fact: if the first occurrence of a value in a list is the same node as the last occurrence of that value, then you know it is unique. As your list is doubly linked, you can use a backwards traversal to find the last occurrence (and a forward traversal to find the first occurrence).
Here is some suggested code:
struct node* findFirstNode(char data) {
struct node *current = head;
while (current != NULL && current->data != data) {
current = current->next;
}
return current;
}
struct node* findLastNode(char data) {
struct node *current = tail;
while (current != NULL && current->data != data) {
current = current->prev;
}
return current;
}
void removeNode(struct node *current) {
if (current->prev == NULL) {
head = current->next;
} else {
current->prev->next = current->next;
}
if (current->next == NULL) {
tail = current->prev;
} else {
current->next->prev = current->prev;
}
free(current);
}
void remove_unique_dll() {
struct node *current = head;
struct node *next;
while (current != NULL)
{
next = current->next;
if (findFirstNode(current->data) == findLastNode(current->data)) {
removeNode(current);
}
current = next;
}
}
You have at least three errors.
After counting the number of occurrences of an item, you use next in several places. However, next has been used to iterate through the list. It was moved to the end and is now a null pointer. You can either reset it with next = current->next; or you can change the places that use next to current->next.
At the end of remove_unique_dll, you have head=current;. There is no reason to update head at this point. Whenever the first node was removed from the list, earlier code in remove_unique_dll updated head. So it is already updated. Delete the line head=current;.
That will leave code that deletes all but one occurrence of each item. However, based on your sample output, you want to leave multiple occurrences of items for which there are multiple occurrences. For that, you need to rethink your logic in remove_unique_dll about deciding which nodes to delete. When it sees the first a, it scans the remainder of the list and sees the second, so it does not delete the first a. When it sees the second a, it scans the remainder of the list and does not see a duplicate, so it deletes the second a. You need to change that.
Let's consider your code step by step.
It seems you think that in this declaration
struct node *head, *tail = NULL; //Represent the head and tail of the doubly linked list
the both pointers head and tail are explicitly initialized by NULL. Actually only the pointer tail is explicitly initialized by NULL. The pointer head is initialized implicitly as a null pointer only due to placing the declaration in file scope. It to place such a declaration in a block scope then the pointer head will be uninitialized.
Instead you should write
struct node *head = NULL, *tail = NULL; //Represent the head and tail of the doubly linked list
Also it is a very bad approach when the functions depend on these global variables. In this case you will be unable to have more than one list in a program.
Also the declaration of the variable len that is used only in main as a global variable
int len;
also a bad idea. And moreover this declaration is redundant.
You need to define one more structure that will contain pointers head and tail as data members as for example
struct list
{
struct node *head;
struct node *tail;
};
The function addNode can invoke undefined behavior when a new node can not be allocated
void addNode(char data)
{
struct node *newNode = (struct node*) malloc(sizeof(struct node)); //Create new node
//...
You should check whether a node is allocated successfully and only in this case change its data members. And you should report the caller whether a node is created or not.
So the function should return an integer that will report an success or failure.
In the function remove_unique_dll after this while loop
while (next != NULL)
{
if (next->data == current->data)
{
cnt += 1;
next = next->next;
}
else
next = next->next;
//printf("#%c %d %c\n",next->data,cnt,current->data);
}
if cnt is equal to 1
if (cnt == 1)
//..
then the pointer next is equal to NULL. And using the pointer next after that like
if (prev == NULL)
{
head = next;
}
else
{
prev->next = next;
}
is wrong.
Also you need to check whether there is a preceding node with the same value as the value of the current node. Otherwise you can remove a node that is not a unique because after it there are no nodes with the same value.
And this statement
head = current;
does not make sense because after the outer while loop
while (current != NULL)
the pointer current is equal to NULL.
Pay attention that the function will be more useful for users if it will return the number of removed unique elements.
Here is a demonstration program that shows how the list and the function remove_unique_dll can be defined.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
struct node
{
char data;
struct node *next;
struct node *prev;
};
struct list
{
struct node *head;
struct node *tail;
};
int addNode( struct list *list, char data )
{
struct node *node = malloc( sizeof( *node ) );
int success = node != NULL;
if (success)
{
node->data = data;
node->next = NULL;
node->prev = list->tail;
if (list->head == NULL)
{
list->head = node;
}
else
{
list->tail->next = node;
}
list->tail = node;
}
return success;
}
size_t remove_unique_dll( struct list *list )
{
size_t removed = 0;
for ( struct node *current = list->head; current != NULL; )
{
struct node *prev = current->prev;
while (prev != NULL && prev->data != current->data)
{
prev = prev->prev;
}
if (prev == NULL)
{
// there is no preceding node with the same value
// so the current node is possibly unique
struct node *next = current->next;
while (next != NULL && next->data != current->data)
{
next = next->next;
}
if (next == NULL)
{
// the current node is indeed unique
struct node *to_delete = current;
if (current->prev != NULL)
{
current->prev->next = current->next;
}
else
{
list->head = current->next;
}
if (current->next != NULL)
{
current->next->prev = current->prev;
}
else
{
list->tail = current->prev;
}
current = current->next;
free( to_delete );
++removed;
}
else
{
current = current->next;
}
}
else
{
current = current->next;
}
}
return removed;
}
void display( const struct list *list )
{
for (const node *current = list->head; current != NULL; current = current->next)
{
printf( "%c<->", current->data );
}
puts( "null" );
}
int main()
{
struct list list = { .head = NULL, .tail = NULL };
const char *s = "aabc";
for (const char *p = s; *p != '\0'; ++p)
{
addNode( &list, *p );
}
printf( "Doubly linked list:\n" );
display( &list );
size_t removed = remove_unique_dll( &list );
printf( "There are removed %zu unique value(s) in the list.\n", removed );
printf( "Doubly linked list after removing unique elements:\n" );
display( &list );
}
The program output is
Doubly linked list:
a<->a<->b<->c<->null
There are removed 2 unique value(s) in the list.
Doubly linked list after removing unique elements:
a<->a<->null
You will need at least to write one more function that will free all the allocated memory when the list will not be required any more.

Swapping neighbor linked list nodes

I have to make a function which swaps the neighbor nodes in a linked list with sentinel. Something like this: 1-2-3-4-5 -> 2-1-4-3-5, but I don't know how to do that. Can somebody help me?
#include <stdio.h>
#include <stdlib.h>
typedef struct _listelem {
int a;
struct _listelem* next;
} listelem;
void reverse_pairs(listelem* a)
{
listelem* head = NULL;
listelem* tail = NULL;
head = a->next;
tail = a->next;
while (head->next != NULL)
{
head = head->next->next;
tail = head;
}
return head;
}
You did not show how the list with a sentinel node is built.
I suppose that the sentinel node is the first node in the list pointed to by the pointer head.
In this case the function can look the following way.
void reverse_pairs( listelem *head )
{
if (head)
{
for (; head->next && head->next->next; head = head->next->next)
{
listelem *tmp = head->next;
head->next = head->next->next;
tmp->next = head->next->next;
head->next->next = tmp;
}
}
}
As for your function implementation then it is incorrect at least because a function with the return type void may not have a statement like this
return head;
Also within this while loop
while (head->next != NULL)
{
head = head->next->next;
tail = head;
}
you are changing the local variables head and tail. Such changes do not influence on the original list.
If you have a circular list when the data member next of the last node points to the head (sentinel) node then the function can look the following way.
void reverse_pairs( listelem *head )
{
if (head)
{
for ( listelem *current = head;
current->next != head && current->next->next != head;
current = current->next->next)
{
listelem *tmp = current->next;
current->next = current->next->next;
tmp->next = current->next->next;
current->next->next = tmp;
}
}
}
While the answer from #VladFromMoscow shows the proper approach for swapping nodes in the list to accomplish your objective, if you are stuck passing a single pointer, and the function return type is fixed at void, then there is another way to go about it.
Instead of swapping nodes, you simply swap the int member value between nodes. Approaching the problem that way, the address of the first node never changes, so there is no need to pass the address of the list as a parameter.
The approach is simple. Take the current node, swap the integer value between the current and next node and advance past the nodes holding the swapped integers. To do so, you advance from the current node to the next and check if the node that follows next is NULL (marking the end of the list). If it is NULL, you are done. If it is not NULL, advance again and repeat. You can write the function either with a while() loop, e.g.
void reverse_pairs (listelem *head)
{
while (head && head->next) {
int tmp = head->a;
head->a = head->next->a;
head->next->a = tmp;
head = head->next;
if (head->next)
head = head->next;
}
}
Or, slightly less readable, using a for() loop and a ternary, e.g.
void reverse_pairs (listelem *head)
{
for (; head && head->next;
head = head->next->next ? head->next->next : NULL) {
int tmp = head->a;
head->a = head->next->a;
head->next->a = tmp;
}
}
Example Use/Output
With a normal list where the last pointer is NULL, your output printing the original list, calling reverse_pairs(), and then outputting the modified list would look as follows:
$ ./bin/lls_revpairs
1 2 3 4 5
2 1 4 3 5
Complete Test Code
The complete test code is included below. Compiling as normal will use the for() loop above, or adding the define -DUSEWHILE, to your compile string will cause the while() loop form of the function to be used:
#include <stdio.h>
#include <stdlib.h>
typedef struct _listelem {
int a;
struct _listelem* next;
} listelem;
/** add node at end of list, update tail to end */
listelem *add (listelem **head, int v)
{
listelem **ppn = head, /* pointer to pointer to head */
*pn = *head, /* pointer to head */
*node = malloc (sizeof *node); /* allocate new node */
if (!node) { /* validate allocation */
perror ("malloc-node");
return NULL;
}
node->a = v; /* initialize members values */
node->next = NULL;
while (pn) {
ppn = &pn->next;
pn = pn->next;
}
return *ppn = node; /* add & return new node */
}
#ifdef USEWHILE
/** reverse node pairs in list - while loop */
void reverse_pairs (listelem *head)
{
while (head && head->next) {
int tmp = head->a;
head->a = head->next->a;
head->next->a = tmp;
head = head->next;
if (head->next)
head = head->next;
}
}
#else
/** reverse node pairs in list - for loop + ternary */
void reverse_pairs (listelem *head)
{
for (; head && head->next;
head = head->next->next ? head->next->next : NULL) {
int tmp = head->a;
head->a = head->next->a;
head->next->a = tmp;
}
}
#endif
/** print all nodes in list */
void prn_list (listelem *l)
{
if (!l) {
puts ("list-empty");
return;
}
size_t i = 0;
for (listelem *n = l; n && i < 10; n = n->next, i++)
printf (" %d", n->a);
putchar ('\n');
}
int main (void) {
listelem *list = NULL;
for (int i = 1; i <= 5; i++)
add (&list, i);
prn_list (list);
reverse_pairs (list);
prn_list (list);
}

How can I fix this error in c code for linked lists?

I am receiving an error on Xcode when for my delete function for a single linked list in c and at this point im not sure why. The error im receiving is
"malloc: * error for object 0x7ffeefbff5d8: pointer being freed was not allocated * set a breakpoint in malloc_error_break to debug"
int delete (node ** list, int delete)
{
node * current= *list;//declares current
//stop at node that is one before i will need to delete (in this case iam deleting the 2node)
if (*list != NULL)
{
node * temp= *list;
temp = temp->next;
free(list);
*list=temp;
}
while (current->next->data != delete)
{
current=current->next;
}
node * temp;//Declares temp
temp=current->next->next;
free(current->next);//one after current free'd
current->next=temp;
return 0;
}
There are number of things which is not clear or unnecessary
Why you want to free the free(list); list is double pointer, if you free this headpointer of linked list will loose. after delete operation How will print/access the list again if needed ?
list is the headpointer which is holding the link and you didn't allocated any memory for list in main() or calling function so you can't free it. thats why error is
pointer being freed was not allocated
If you want to free only then first you should do free(*list) and then free(list)
From your code segment
free(list); /* list is freed here */
*list=temp; /* How will you access again *list */
As per my thinking your if block should be
if (*list != NULL) {
node * temp= *list;
temp = temp->next; /* temp holds 2nd node address & you need that only */
}
First of all Your question was very unclear, for the next time please post all of the necessary code and explain the purpose of the function in a few line.
I guess what you are trying to do is to remove a specific item from a linked list according to the value of the parameter data.
From the site learn-c:
"
Removing a specific item
To remove a specific item from the list, either by its index from the beginning of the list or by its value, we will need to go over all the items, continuously looking ahead to find out if we've reached the node before the item we wish to remove. This is because we need to change the location to where the previous node points to as well.
Here is the algorithm:
Iterate to the node before the node we wish to delete
Save the node we wish to delete in a temporary pointer
Set the previous node's next pointer to point to the node after the node we wish to delete
Delete the node using the temporary pointer
There are a few edge cases we need to take care of, so make sure you understand the code. "
Then they posted the code, I think you should go there and read the entire post.
Good luck
Your code have a faulty logic plus obvious errors.
Once list is freed you have no right to access it again.
free(list);
*list=temp;
Managing the linked list is not difficult but it requires attention to details.
When you delete the node in the list you have a responsibility to connect the nodes.
If you delete the head you have a responsibility to move the head.
If the head is the node you are looking for and this is the only node in the list than after removal that node has to be marked as NULL.
The test program below uses struct node *find(struct node *start, int data) function to find a node which matches your criteria and uses delete to delete it. All edge cases have been taken care of.
#include <stdio.h>
#include <stdlib.h>
// Basic simple single list implementation to illustrate
// a proper deletion of the node which has a specfic data value.
// Node in List
typedef struct node {
int data;
struct node* next; // pointer to next node
}node;
// List structure
typedef struct list {
node* head; // The entry point into a linked list. If the list is empty then the head is a null reference.
} list;
// Create list
list* list_create()
{
list* new_list = malloc(sizeof(list));
if(new_list == NULL)
return NULL; // protection
new_list->head = NULL; // If the list is empty then the head is a null reference. no elements in the list
return new_list; // return created new list
}
// returns newly created node
node* node_new(int data)
{
node* new_node = malloc(sizeof(node)); // allocate memory for the node
if (new_node == NULL)
return NULL; // protection
new_node->data = data; // remember the data
new_node->next = NULL; // no next node
return new_node; // return new created node
}
// The method creates a node and prepends it at the beginning of the list.
//
// Frequently used names for this method:
//
// insert at head
// add first
// prepend
//
// returns new head or NULL on failer
node* prepend_node(list* in_list, node* new_node)
{
// Add item to the front of the in_list, return pointer to the prepended node (head)
if(in_list == NULL)
return NULL;
if(new_node == NULL) // problem, not enough memory
return NULL; // in_list->head has not changed
/*
new_node
|*| --> NULL
next
*/
if(in_list->head == NULL) // if list is empty
{
in_list->head = new_node; // the new_node becomes a head
}
else // list already have a head node
{
/*
|2|-->|1|-->NULL
^
|
*
head (2) (list pointer)
*/
new_node->next = in_list->head; // now, the new node next pointer points to the node pointed by the list head, see below:
/*
new_node
|3|--> |2|-->|1|-->NULL
^
|
*
head (list pointer)
*/
in_list->head = new_node; // the list head has to move to new_node ( a new prepanded node)
/*
new_node
|3|--> |2|-->|1|-->NULL
^
|
*
head (3) (list pointer)
*/
}
return in_list->head; // we are returning pinter to new_node
}
// Print out list
void print_list(list* in_list)
{
node* node;
if (in_list == NULL)
{
return;
}
if (in_list->head == NULL)
{
printf("List is empty!\n");
return;
}
printf("List: ");
node = in_list->head;
while(node != NULL)
{
printf(" %d", node->data);
node = node->next;
}
printf("\n");
}
struct node *find(struct node *start, int data) // find p to be removed
{
node* node;
if (start == NULL)
return NULL;
node = start;
while(node != NULL)
{
if (node->data == data)
return node;
node = node->next;
}
return NULL;
}
int delete(struct node **start, int data)
{
struct node *p, *prev, *next, *to_free;
if (start == NULL) // protection
return 0;
p = find(*start, data); // find p to be removed
if (p == NULL)
return 0;
if (*start == NULL)
return 0; // protection
if(*start == p) // head == p
{
if((*start)->next !=NULL)
{
*start = (*start)->next; // remember next
free(p);
printf("Head removed\n");
return 1;
}
else // the only node
{
free(p);
printf("Last node removed\n");
*start = NULL;
return 1;
}
}
// p != start:
next = *start;
while (next != NULL)
{
prev = next;
to_free = next->next; // candidate to be freed
if( to_free == p )
{
prev->next = to_free->next; // connect nodes before deletion
free(to_free); // now free the remembered `next`
to_free = NULL; // so it does not point to the released memory
return 1;
}
next = next->next; // this node was not a match
} //while
return 0;
}
int main() {
list* new_list = list_create();
node *n1 = node_new(1);
node *n2 = node_new(2);
node *n3 = node_new(3);
// list is empty
print_list(new_list);
prepend_node(new_list, n1);
prepend_node(new_list, n2);
prepend_node(new_list, n3);
// list has 3 elements
print_list(new_list);
delete(&new_list->head, 3);
print_list(new_list);
delete(&new_list->head, 1);
print_list(new_list);
delete(&new_list->head, 2);
// list has 2 elements
print_list(new_list);
printf("head: %p\n",new_list->head);
print_list(new_list);
free (new_list); // after deleting all elements, delete the list itself
return 0;
}
Output:
List is empty!
List: 3 2 1
Head removed
List: 2 1
List: 2
Last node removed
List is empty!
head: (nil)
List is empty!

Why position 0 in delete node does not work?

Im new to c programming. I wanted to create a linked list from a given file and then randomly get a node from linked list then delete that node.
So the code works great but for the position 0 in linked list does not work.
Please help me
here's the code:
typedef struct node{
int *name;
struct node *next;
}node;
delete node:
void deleteNode(node **head_ref, int position){
if(*head_ref == NULL){
return;
}
node * temp = *head_ref;
if(position == 0)
{
*head_ref = (*head_ref)->next;
return;
}
int h;
for(h=0 ; temp!=NULL && h<position-1 ; h++){
temp = temp->next;
}
if(temp == NULL || temp->next == NULL)
return;
node * next = temp->next->next;
free(temp->next);
temp->next = next;}
getting random node:
void RandomFromList(node *head){
// IF list is empty
if (head == NULL){
return -1;
}
word = head->name;
// Iterate from the (k+1)th element to nth element
node *current = head;
int n;
for (n=2; current!=NULL; n++)
{
// change result with probability 1/n
if (rand() % n == 0)
word = current->name;
// Move to next node
current = current->next;
}
sprintf(words , "%s" , word);
deleteNode(&head , search(head , word));
printf("Randomly selected key is %s\n", words);}
and the file Reader:
node* fileReader(FILE *file){
node *t = malloc(sizeof(node));
char TopicName[20];
int fileRead = fscanf(file,"%s",TopicName);
if(fileRead != EOF){
t->name = strdup(TopicName);
tedad++;
t->next = fileReader(file);
}
if(fileRead == EOF) {
return NULL;
}
return t;}
EDIT:
When the code run's and when the position randomly got 0 the 0 position of linked list doesn't delete and continues with that node in linked list.
EDIT2:I changed my delete node and it works well without any problem, thank you guys!
node* deleteNode(node* head, unsigned i){
node* next;
if(head == NULL)
return head;
next = head->next;
return i == 0
? (free(head), next)
: (head->next = delete_at_index(next, i - 1), head);
}
The major logical problem I see with your delete function is that it is void, i.e. it returns nothing. This is fine if the node being deleted is in the middle (or end) of the list, because the head does not change. But for the case of deleting the head, the caller might expect that his reference would then point to the next node (or null, if a list of one element) after making the call. Consider this code:
node* deleteNode (node *head_ref, int position)
{
// passing in a NULL list returns NULL
if (head_ref == NULL) {
{
return NULL;
}
// deleting the first element returns the second element as the new head
node* temp = head_ref;
if (position == 0)
{
node* ret = temp->next;
free(head_ref);
return ret;
}
// otherwise walk down the list to one before the deletion position
for (int h=0; temp != NULL && h < position-1; h++) {
temp = temp->next;
}
// if found, delete the node at the desired position
if (temp != NULL && temp->next == NULL) {
node* next = temp->next->next;
free(temp->next);
temp->next = next;
}
// for cases other than deleting the head, just return the current
// (unmodified) head
return head_ref;
}
This isn't related to your problem, but don't forget to free the memory:
node * temp = *head_ref;
if(position == 0)
{
*head_ref = temp->next;
free(temp); // <--------
return;
}
Also, you already have a pointer (temp) to *head_ref, it looks cleaner to me to just use that pointer instead of dereferencing head_ref again.
void deleteNode(node **head_ref, int pos){
node *del;
for ( ; *head_ref; head_ref = &(*head_ref)->next) {
if (pos-- <= 0) break;
}
if (!*head_ref) return; // Reached end of list: nothing found
del = *head_ref;
*head_ref = del->next;
free(del);
return;
}
If you want to keep deleteNode void, then the problem is with your RandomFromList function. You are just changing the * head that exists in the function body not the pointer you passed to the function, so it's still pointing to the previous node.
It's because that pointers are passed by value (copied) like other things in C.
Try making RandomFromList return the head pointer.
P.s. I think you also have some memory leaks in the delete function.

Inserting to head and tail of a doubly linked list -- only prints out last tail item inserted

I am trying to implement a doubly linked list in order to study for an exam and am running into some trouble when inserting items into the tail -- it prints out correctly when I only insert to the head in the list. However, when I insert to the tail in the list only the last tail item is printed out. Below is my full code:
/*
- Q2: Write a function that takes a LinkedList struct pointer and inserts at the head of the linked list.
- Q3. Write a function that takes a LinkedList struct pointer and frees all memory associated with it. (
For a solution, see fancy-linked-lists.c, attached above.)
- Q4 Review all the functions from today's code and give their big-oh runtimes.
- Q5: Implement functions for doubly linked lists:
- tail_insert(),
- head_insert(),
- tail_delete(),
- head_delete(),
- delete_Nth().
- Repeat these exercises with doubly linked lists in which you maintain a tail pointer.
- How does the tail pointer affect the runtimes of these functions?
- Are any of these functions more efficient for doubly linked lists with tail pointers than they are for singly linked lists with tail pointers?
*/
#include <stdio.h>
#include <stdlib.h>
typedef struct node
{
int data;
struct node *next;
struct node *prev;
} node;
node *createNode(int data)
{
node *ptr = NULL;
ptr = malloc(sizeof(node));
if(ptr == NULL)
{
printf("space could not be allocated\n");
return NULL;
}
ptr->data = data;
ptr->next = NULL;
ptr->prev = NULL;
return ptr;
}
node *tailInsert(node *head, int data)
{
if(head->next == NULL)
{
node *temp;
temp = createNode(data);
temp->next = NULL;
temp->prev = head;
head->next = temp;
return head;
}
tailInsert(head->next, data);
}
node *frontInsert(node *head, int data)
{
node *newHead;
if(head == NULL)
{
return createNode(data);
}
newHead = createNode(data);
newHead->next = head;
newHead->prev = NULL;
return newHead;
}
node *destroy_linked_list(node *list)
{
/*if (list == NULL)
return NULL;
// Free the entire list within this struct.
destroy_list(list->head);
// Free the struct itself.
free(list);
return NULL;
*/
}
void printList(node *head)
{
if (head == NULL)
{
printf("Empty List\n");
return;
}
for(; head != NULL; head = head->next)
printf("%d ", head->data);
printf("\n");
}
int main(void)
{
node *head = NULL;
head = frontInsert(head, 1);
head = frontInsert(head, 2);
head = frontInsert(head, 3);
head = tailInsert(head, 4);
head = tailInsert(head, 5);
head = tailInsert(head, 6);
printList(head);
system("PAUSE");
return 0;
}
I appreciate the help!
You are returning the temp(last node) from tailInsert and assigning to head. Do not change the head pointer if you are inserting at tail.
void tailInsert(node *head, int data)
{
if(head->next == NULL)
{
node *temp;
temp = createNode(data);
temp->next = NULL;
temp->prev = head;
head->next = temp;
return ;
}
tailInsert(head->next, data);
}
In main do not assign anything to head if you are calling tailInsert function

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