# 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 below. 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. We will subclass UnorderedList and leave the __init__ method intact as once again, an empty list will be denoted by a head reference to None.

from unordered_list import Node, UnorderedList

class OrderedList(UnorderedList):


As we consider the operations for the ordered list, we should note that the is_empty and 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 methods, search and 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, the diagram below 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.

Below we provide an adaptation of the search method from our UnorderedList class to take advantage of this optimization.

    def search(self, item):

while current is not None:
if current.value == item:
return True
if current.value > item:
return False
current = current.next

return False


The most significant method modification will take place in add. 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. Below we show 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 (current becomes 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 to stop.

As we saw with unordered lists, it is necessary to have an additional reference, again called previous, since current will not provide access to the node that must be modified.

Once we have identified the position at which to add our new node, we construct it and place it correctly, either as the new head of the node (if previous is None) or between previous and current otherwise.

    def add(self, item):
previous = None

while current is not None:
if current.value > item:
break
previous, current = current, current.next

temp = Node(item)
if previous is None:
else:
temp.next, previous.next = current, temp


We leave the remaining methods as exercises. You should carefully consider whether the unordered implementations will work given that the list is now ordered.

To analyze the complexity of the linked list operations, we need to consider whether they require traversal. Consider a linked list that has n nodes. The is_empty method is $O(1)$ since it requires one step to check the head reference for None. size, on the other hand, will always require $n$ steps since there is no way to know how many nodes are in the linked list without traversing from head to end. Therefore, length is $O(n)$. Adding an item to an unordered list will always be $O(1)$ since we simply place the new node at the head of the linked list. However, search and remove, as well as add for an ordered list, all require the traversal process. Although on average they may need to traverse only half of the nodes, these methods are all $O(n)$ since in the worst case each will process every node in the list.

You may also have noticed that the performance of this implementation differs from the actual performance given earlier for Python lists. This suggests that linked lists are not the way Python lists are implemented. The actual implementation of a Python list is based on the notion of an array. We discuss this in depth later.

# Practical Algorithms and Data Structures

Introduction

## Lists

1. Introduction to Lists
2. Implementing an Unordered List
3. Implementing an Ordered List