How do I delete a node from linked list? - c++

Question

How can I delete a node (between two nodes) from a single linked list without passing any parameters to the class function?
For example, I have a list of 6 nodes with one head node and I want to delete two of them (without prior knowledge of their address or position) from a class function, how would I do that?
void WordList::deleteNode(){
Node *temp;
temp=head;
if(temp->count<=10)
{
//delete this node... not sure how though
}
else
temp=temp->next;
}
where WordList is my class, Node is my struct which holds a word, a count, and a pointer.
I want to delete any node that has a counter of 10 or less.

Answer 1

Your edit has prior information, the bit that states "counter <= 10" :-)
Pseudo-code for deleting elements meeting that criteria in a singly-linked list:
def delLessThanTen:
# Delete heads meeting criteria, stop when list empty.
while head != NULL and head->count <= 10:
temp = head->next
free head
head = temp
if head == NULL:
return
# Head exists, with count > 10, process starting there (we check
# NEXT element for criteria then delete if met).
ptr = head
while ptr->next != NULL:
# If next in list meets criteria, delete it, otherwise advance.
if ptr->next->count <= 10:
temp = ptr->next->next
free ptr->next
ptr->next = temp
else:
ptr = ptr->next
return

Answer 2

I find the question too confusing.
Deletion of a node from the list is always based on some criteria e.g. the content of the element, the position of the element etc (unless you are deleting all the elements in the list)

Answer 3

something like this:
void WordList::deleteNode(){
Node *prev=NULL;
temp=head;
bool done=false;
while (!done)
{
if (temp->count<=10)
{
if (prev == NULL)
{
head = temp->next;
} else
{
prev->next = temp->next;
}
// delete data in temp, and the node if necessary
temp = temp->next;
done = (temp==NULL) || // some other condition, like deleted 2
} else
{
prev=temp;
temp = temp->next;
done = (temp==NULL);
}
}

Answer 4

Have a previous variable initialized to null. If you delete a node, change previous's next to the element's next, unless previous is null (you are at the start of the list) when you leave previous null and change root to the deleted element's next. If you don't delete the element, change previous to the element.
Here previous will always point to the previous element or be null if you're at the start of the list.
void WordList::deleteNode() {
Node *temp = head;
Node *previous = null;
while (temp != null) {
if(temp->count <= 10) {
// delete node
if (previous == null) {
// there is no previous node, so point head of list past the current node
head = temp->next;
} else {
// there is a previous node, so just point it past the current node
previous->next = temp->next;
}
} else {
// not deleting, so set previous to temp
previous = temp;
}
temp = temp->next;
}
}