4.23. Implementing an Ordered List¶
In order to implement the ordered list, we must remember that the relative positions of the items are based on some underlying characteristic. The ordered list of integers given above (17, 26, 31, 54, 77, and 93) can be represented by a linked structure as shown in Figure 15. Again, the node and link structure is ideal for representing the relative positioning of the items.
To implement the
OrderedList class, we will use the same technique
as seen previously with unordered lists. Once again, an empty list will
be denoted by a
head reference to
class OrderedList: def __init__(self): self.head = None
As we consider the operations for the ordered list, we should note that
size methods can be implemented the same as
with unordered lists since they deal only with the number of nodes in
the list without regard to the actual item values. Likewise, the
remove method will work just fine since we still need to find the
item and then link around the node to remove it. The two remaining
add, will require some modification.
The search of an unordered linked list required that we traverse the
nodes one at a time until we either find the item we are looking for or
run out of nodes (
None). It turns out that the same approach would
actually work with the ordered list and in fact in the case where we
find the item it is exactly what we need. However, in the case where the
item is not in the list, we can take advantage of the ordering to stop
the search as soon as possible.
For example, Figure 16 shows the ordered linked list as a
search is looking for the value 45. As we traverse, starting at the head
of the list, we first compare against 17. Since 17 is not the item we
are looking for, we move to the next node, in this case 26. Again, this
is not what we want, so we move on to 31 and then on to 54. Now, at this
point, something is different. Since 54 is not the item we are looking
for, our former strategy would be to move forward. However, due to the
fact that this is an ordered list, that will not be necessary. Once the
value in the node becomes greater than the item we are searching for,
the search can stop and return
False. There is no way the item could
exist further out in the linked list.
Listing 9 shows the complete
search method. It is
easy to incorporate the new condition discussed above by adding another
stop, and initializing it to
False (line 4).
stop) we can continue to look
forward in the list (line 5). If any node is ever discovered that
contains data greater than the item we are looking for, we will set
True (lines 9–10). The remaining lines are identical to
the unordered list search.
def search(self,item): current = self.head found = False stop = False while current != None and not found and not stop: if current.getData() == item: found = True else: if current.getData() > item: stop = True else: current = current.getNext() return found
The most significant method modification will take place in
Recall that for unordered lists, the
add method could simply place a
new node at the head of the list. It was the easiest point of access.
Unfortunately, this will no longer work with ordered lists. It is now
necessary that we discover the specific place where a new item belongs
in the existing ordered list.
Assume we have the ordered list consisting of 17, 26, 54, 77, and 93 and
we want to add the value 31. The
add method must decide that the new
item belongs between 26 and 54. Figure 17 shows the setup
that we need. As we explained earlier, we need to traverse the linked
list looking for the place where the new node will be added. We know we
have found that place when either we run out of nodes (
None) or the value of the current node becomes greater than
the item we wish to add. In our example, seeing the value 54 causes us
As we saw with unordered lists, it is necessary to have an additional
reference, again called
current will not provide
access to the node that must be modified. Listing 10 shows
add method. Lines 2–3 set up the two external
references and lines 9–10 again allow
previous to follow one node
current every time through the iteration. The condition (line
5) allows the iteration to continue as long as there are more nodes and
the value in the current node is not larger than the item. In either
case, when the iteration fails, we have found the location for the new
The remainder of the method completes the two-step process shown in
Figure 17. Once a new node has been created for the item,
the only remaining question is whether the new node will be added at the
beginning of the linked list or some place in the middle. Again,
previous == None (line 13) can be used to provide the answer.
def add(self,item): current = self.head previous = None stop = False while current != None and not stop: if current.getData() > item: stop = True else: previous = current current = current.getNext() temp = Node(item) if previous == None: temp.setNext(self.head) self.head = temp else: temp.setNext(current) previous.setNext(temp)
OrderedList class with methods discussed thus far can be found
in ActiveCode 1.
We leave the remaining methods as exercises. You should carefully
consider whether the unordered implementations will work given that the
list is now ordered.