- Another popular way to insert elements into a linked list is
ordered by the value that is inserted.
- These notes show you have you what you must do to maintain the elements in a linked list in ascending order... However, I do not provide any program code. You have to figure out the details using what you have learned in a homework project.
- Pictorial representation of the BEFORE and AFTER states:
- Recipe to insert a new element into an ordered list
(general case --- NOT the exception cases):
- Find adjacent elements
p1 and
p2 in the list such that:
- p1->value
is
smaller than or equal to the new element
- p2->value is greater than or equal to the value in the new element
- p1->value
is
smaller than or equal to the new element
- Make
p1->next
point to the new element
(the new element will thus "follow" this element)
- Set the "next" linkage variable in the new element to point to p2 (the element p2 follows the new element)
Pictorial representation of the insert algorithm:
- Find adjacent elements
p1 and
p2 in the list such that:
- There are 2 exception cases
that must be handled
differently when inserting
in an ordered list:
- When the list is empty
- When the new element that is being inserted into the list will become the first element of the list
All other cases can be handle by the insertion algorithm described above.
- When the list is empty
- SPECIAL CASE 1:
inserting in
an EMPTY list:
- How to detect this case:
-
head == NULL
- How to handle insertion in this special case:
-
handle it in the same way as in the "insert at tail" case
- How to detect this case:
- SPECIAL CASE 2:
new element occupies the
head position in
an non-empty list:
- How to detect this case:
-
newElem->value <= head->value
- How to handle insertion in this special case:
-
Handle it in the same way we have handle insertion at the head of
a list.
- How to detect this case:
- Program skeleton
if ( .... special case 1 detected ... ) { code to handle insert for special case 1 } else if ( .... special case 2 detected ... ) { code to handle insert for special case 2 } else { code to handle insert for all other cases }