I am creating a linked list as in the previous question I asked. I have found that the best way to develop the linked list is to have the head and tail in another structure.
A demo for Singly Linked List. If you prefer, try to check Circular Linked List and Doubly Linked List.
#include
#include
typedef struct node {
int val;
struct node * next;
} node_t;
// Iterating over a list
void
print_list(node_t *head)
{
node_t *current = head;
while(current != NULL)
{
printf("%d\n", current->val);
current = current->next;
}
}
// Adding an item to the end of the list
void
push_end(node_t *head, int val)
{
node_t *current = head;
while (current->next != NULL)
{
current = current->next;
}
current->next = malloc(sizeof(node_t));
current->next->val = val;
current->next->next = NULL;
}
// Adding an item to the head of the list
void
push_head(node_t **head, int val)
{
node_t *new_node = NULL;
new_node = malloc(sizeof(node_t));
new_node->val = val;
new_node->next = *head;
*head = new_node;
}
// Removing the head item of the list
int
pop_head(node_t **head)
{
int retval = -1;
node_t *next_node = NULL;
if (*head == NULL) {
return -1;
}
next_node = (*head)->next;
retval = (*head)->val;
free(*head);
*head = next_node;
return retval;
}
// Removing the last item of the list
int
pop_last(node_t *head)
{
int retval = 0;
node_t *current = NULL;
if (head->next == NULL) {
retval = head->val;
free(head);
return retval;
}
/* get to the second to last node in the list */
current = head;
while (current->next->next != NULL) {
current = current->next;
}
/* now current points to the second to last item of the list.
so let's remove current->next */
retval = current->next->val;
free(current->next);
current->next = NULL;
return retval;
}
// Removing a specific item
int
remove_by_index(node_t **head, int n)
{
int i = 0;
int retval = -1;
node_t *current = *head;
node_t *temp_node = NULL;
if (n == 0) {
return pop_head(head);
}
for (i = 0; i < n - 1; i++) {
if (current->next == NULL) {
return -1;
}
current = current->next;
}
temp_node = current->next;
retval = temp_node->val;
current->next = temp_node->next;
free(temp_node);
return retval;
}
int
main(int argc, const char *argv[])
{
int i;
node_t * testnode;
for (i = 0; i < argc; i++)
{
push_head(&testnode, atoi(argv[i]));
}
print_list(testnode);
return 0;
}
// http://www.learn-c.org/en/Linked_lists
// https://www.geeksforgeeks.org/data-structures/linked-list/