How to delete a node in a singly link list with only one pointer pointing to node to be deleted?
[Start and end pointers are not known, the available information is poin
Not possible.
There are hacks to mimic the deletion.
But none of then will actually delete the node the pointer is pointing to.
The popular solution of deleting the following node and copying its contents to the actual node to be deleted has side-effects if you have external pointers pointing to nodes in the list, in which case an external pointer pointing to the following node will become dangling.
You can find some discussion on SO here.
The only sensible and safe option under such restrictions is to mark the node deleted without actually unlinking it, deferring that to a later time.
You can delete a node without getting the previous node, by having it mimic the following node and deleting that one instead:
void delete(Node *n) {
if (!is_sentinel(n->next)) {
n->content = n->next->content;
Node *next = n->next;
n->next = n->next->next;
free(next);
} else {
n->content = NULL;
free(n->next);
n->next = NULL;
}
}
As you can see, you will need to deal specially for the last element. I'm using a special node as a sentinel node to mark the ending which has content
and next
be NULL
.
UPDATE: the lines Node *next = n->next; n->next = n->next->next
basically shuffles the node content, and frees the node: Image that you get a reference to node B to be deleted in:
A / To be deleted
next ---> B
next ---> C
next ---> *sentinel*
The first step is n->content = n->next->content
: copy the content of the following node to the node to be "deleted":
A / To be deleted
next ---> C
next ---> C
next ---> *sentinel*
Then, modify the next
points:
A / To be deleted
next ---> C /----------------
next ---| C |
next ---> *sentinel*
The actually free the following element, getting to the final case:
A / To be deleted
next ---> C
next ---> *sentinel*