问题
I'm stuck on this particular function that frees all even nodes from the linked list. I've figured out how to free all the nodes from a linked list but I cannot figure this out. The code i'm posting is very wrong. What I don't understand is how to use a node *temp variable and link it to the head->next node, as the head is what is being freed (because it is even). Also, at the end of the while loop, I know that I need to increment to the next node in the list, but I seem to be doing that already in the first if statement, so wouldn't calling current = current->next actually be taking me to current->next->next, and skipping a node? Sorry for this massive block of text.
node *delete_even_node(node *head)
{
node *temp, *current = head;
if (head == NULL)
return NULL;
while (current != NULL)
{
if (current->data % 2 == 0)
{
printf("Deleting Even %d\n", current->data);
temp = current->next; //problem starts
temp = temp->next;
free(temp);
}
else
current = current->next;
current = current->next;
}
return head;
}
回答1:
There are two approaches.
If to use your approach when the pointer to the head node is passed by value then the function can look the following way.
node * delete_even_node( node *head )
{
while ( head && head->data % 2 == 0 )
{
node *tmp = head;
head = head->next;
free( tmp );
}
if ( head )
{
node *current = head;
while ( current->next )
{
if ( current->next->data % 2 == 0 )
{
node *tmp = current->next;
current->next = current->next->next;
free( tmp );
}
else
{
current = current->next;
}
}
}
return head;
}
Another approach is to pass the pointer to the head node by reference.
For example
void delete_even_node( node **head )
{
while ( *head )
{
if ( ( *head )->data % 2 == 0 )
{
node *tmp = *head;
*head = ( *head )->next;
free( tmp );
}
else
{
head = &( *head )->next;
}
}
}
And the function is called like
delete_even_node( &head );
来源:https://stackoverflow.com/questions/64527206/deleting-node-inside-of-linked-list