4.21. Implementing an Unordered List: Linked Lists¶
In order to implement an unordered list, we will construct what is commonly known as a linked list. Recall that we need to be sure that we can maintain the relative positioning of the items. However, there is no requirement that we maintain that positioning in contiguous memory. For example, consider the collection of items shown in Figure 1. It appears that these values have been placed randomly. If we can maintain some explicit information in each item, namely the location of the next item (see Figure 2), then the relative position of each item can be expressed by simply following the link from one item to the next.
It is important to note that the location of the first item of the list must be explicitly specified. Once we know where the first item is, the first item can tell us where the second is, and so on. The external reference is often referred to as the head of the list. Similarly, the last item needs to know that there is no next item.
The basic building block for the linked list implementation is the
node. Each node object must hold at least two pieces of information.
First, the node must contain the list item itself. We will call this the
data field of the node. In addition, each node must hold a reference
to the next node. Listing 1 shows the Python
implementation. To construct a node, you need to supply the initial data
value for the node. Evaluating the assignment statement below will yield
a node object containing the value 93 (see Figure 3). You
should note that we will typically represent a node object as shown in
Figure 4. The
Node class also includes the usual methods
to access and modify the data and the next reference.
class Node: def __init__(self,initdata): self.data = initdata self.next = None def getData(self): return self.data def getNext(self): return self.next def setData(self,newdata): self.data = newdata def setNext(self,newnext): self.next = newnext
Node objects in the usual way.
>>> temp = Node(93) >>> temp.getData() 93
The special Python reference value
None will play an important role
Node class and later in the linked list itself. A reference
None will denote the fact that there is no next node. Note in the
constructor that a node is initially created with
next set to
None. Since this is sometimes referred to as “grounding the node,”
we will use the standard ground symbol to denote a reference that is
None. It is always a good idea to explicitly assign
None to your initial next reference values.
Unordered List Class¶
As we suggested above, the unordered list will be built from a
collection of nodes, each linked to the next by explicit references. As
long as we know where to find the first node (containing the first
item), each item after that can be found by successively following the
next links. With this in mind, the
UnorderedList class must maintain
a reference to the first node. Listing 2 shows the
constructor. Note that each list object will maintain a single reference
to the head of the list.
class UnorderedList: def __init__(self): self.head = None
Initially when we construct a list, there are no items. The assignment statement
>>> mylist = UnorderedList()
creates the linked list representation shown in
Figure 5. As we discussed in the
Node class, the
None will again be used to state that the head of
the list does not refer to anything. Eventually, the example list given
earlier will be represented by a linked list as shown in
Figure 6. The head of the list refers to the first node
which contains the first item of the list. In turn, that node holds a
reference to the next node (the next item) and so on. It is very
important to note that the list class itself does not contain any node
objects. Instead it contains a single reference to only the first node
in the linked structure.
isEmpty method, shown in Listing 3, simply checks to
see if the head of the list is a reference to
None. The result of
the boolean expression
self.head==None will only be true if there
are no nodes in the linked list. Since a new list is empty, the
constructor and the check for empty must be consistent with one another.
This shows the advantage to using the reference
None to denote the
“end” of the linked structure. In Python,
None can be compared to
any reference. Two references are equal if they both refer to the same
object. We will use this often in our remaining methods.
def isEmpty(self): return self.head == None
So, how do we get items into our list? We need to implement the
method. However, before we can do that, we need to address the important
question of where in the linked list to place the new item. Since this
list is unordered, the specific location of the new item with respect to
the other items already in the list is not important. The new item can
go anywhere. With that in mind, it makes sense to place the new item in
the easiest location possible.
Recall that the linked list structure provides us with only one entry
point, the head of the list. All of the other nodes can only be reached
by accessing the first node and then following
next links. This
means that the easiest place to add the new node is right at the head,
or beginning, of the list. In other words, we will make the new item the
first item of the list and the existing items will need to be linked to
this new first item so that they follow.
The linked list shown in Figure 6 was built by calling
add method a number of times.
>>> mylist.add(31) >>> mylist.add(77) >>> mylist.add(17) >>> mylist.add(93) >>> mylist.add(26) >>> mylist.add(54)
Note that since 31 is the first item added to the list, it will eventually be the last node on the linked list as every other item is added ahead of it. Also, since 54 is the last item added, it will become the data value in the first node of the linked list.
add method is shown in Listing 4. Each item of the list
must reside in a node object. Line 2 creates a new node and places the
item as its data. Now we must complete the process by linking the new
node into the existing structure. This requires two steps as shown in
Figure 7. Step 1 (line 3) changes the
of the new node to refer to the old first node of the list. Now that the
rest of the list has been properly attached to the new node, we can
modify the head of the list to refer to the new node. The assignment
statement in line 4 sets the head of the list.
The order of the two steps described above is very important. What happens if the order of line 3 and line 4 is reversed? If the modification of the head of the list happens first, the result can be seen in Figure 8. Since the head was the only external reference to the list nodes, all of the original nodes are lost and can no longer be accessed.
def add(self,item): temp = Node(item) temp.setNext(self.head) self.head = temp
The next methods that we will implement–
remove–are all based on a technique known as linked list
traversal. Traversal refers to the process of systematically visiting
each node. To do this we use an external reference that starts at the
first node in the list. As we visit each node, we move the reference to
the next node by “traversing” the next reference.
To implement the
size method, we need to traverse the linked list
and keep a count of the number of nodes that occurred.
Listing 5 shows the Python code for counting the number of
nodes in the list. The external reference is called
current and is
initialized to the head of the list in line 2. At the start of the
process we have not seen any nodes so the count is set to \(0\).
Lines 4–6 actually implement the traversal. As long as the current
reference has not seen the end of the list (
None), we move current
along to the next node via the assignment statement in line 6. Again,
the ability to compare a reference to
None is very useful. Every
time current moves to a new node, we add \(1\) to
count gets returned after the iteration stops.
Figure 9 shows this process as it proceeds down the list.
1def size(self): 2 current = self.head 3 count = 0 4 while current != None: 5 count = count + 1 6 current = current.getNext() 7 8 return count
Searching for a value in a linked list implementation of an unordered list also uses the traversal technique. As we visit each node in the linked list we will ask whether the data stored there matches the item we are looking for. In this case, however, we may not have to traverse all the way to the end of the list. In fact, if we do get to the end of the list, that means that the item we are looking for must not be present. Also, if we do find the item, there is no need to continue.
Listing 6 shows the implementation for the
As in the
size method, the traversal is initialized to start at
the head of the list (line 2). We also use a boolean variable called
found to remember whether we have located the item we are searching
for. Since we have not found the item at the start of the traversal,
found can be set to
False (line 3). The iteration in line 4
takes into account both conditions discussed above. As long as there are
more nodes to visit and we have not found the item we are looking for,
we continue to check the next node. The question in line 5 asks whether
the data item is present in the current node. If so,
found can be
1def search(self,item): 2 current = self.head 3 found = False 4 while current != None and not found: 5 if current.getData() == item: 6 found = True 7 else: 8 current = current.getNext() 9 10 return found
As an example, consider invoking the
search method looking for the
>>> mylist.search(17) True
Since 17 is in the list, the traversal process needs to move only to the
node containing 17. At that point, the variable
found is set to
True and the
while condition will fail, leading to the return
value seen above. This process can be seen in Figure 10.
remove method requires two logical steps. First, we need to
traverse the list looking for the item we want to remove. Once we find
the item (recall that we assume it is present), we must remove it. The
first step is very similar to
search. Starting with an external
reference set to the head of the list, we traverse the links until we
discover the item we are looking for. Since we assume that item is
present, we know that the iteration will stop before
current gets to
None. This means that we can simply use the boolean
found in the
current will be a reference to the
node containing the item to be removed. But how do we remove it? One
possibility would be to replace the value of the item with some marker
that suggests that the item is no longer present. The problem with this
approach is the number of nodes will no longer match the number of
items. It would be much better to remove the item by removing the entire
In order to remove the node containing the item, we need to modify the
link in the previous node so that it refers to the node that comes after
current. Unfortunately, there is no way to go backward in the linked
current refers to the node ahead of the node where we
would like to make the change, it is too late to make the necessary
The solution to this dilemma is to use two external references as we
traverse down the linked list.
current will behave just as it did
before, marking the current location of the traverse. The new reference,
which we will call
previous, will always travel one node behind
current. That way, when
current stops at the node to be removed,
previous will be referring to the proper place in the linked list
for the modification.
Listing 7 shows the complete
remove method. Lines 2–3
assign initial values to the two references. Note that
starts out at the list head as in the other traversal examples.
previous, however, is assumed to always travel one node behind
current. For this reason,
previous starts out with a value of
None since there is no node before the head (see
Figure 11). The boolean variable
found will again be
used to control the iteration.
In lines 6–7 we ask whether the item stored in the current node is the
item we wish to remove. If so,
found can be set to
True. If we
do not find the item,
current must both be moved
one node ahead. Again, the order of these two statements is crucial.
previous must first be moved one node ahead to the location of
current. At that point,
current can be moved. This process is
often referred to as “inch-worming” as
previous must catch up to
current moves ahead. Figure 12 shows
the movement of
current as they progress down the
list looking for the node containing the value 17.
1def remove(self,item): 2 current = self.head 3 previous = None 4 found = False 5 while not found: 6 if current.getData() == item: 7 found = True 8 else: 9 previous = current 10 current = current.getNext() 11 12 if previous == None: 13 self.head = current.getNext() 14 else: 15 previous.setNext(current.getNext())
Once the searching step of the
remove has been completed, we need to
remove the node from the linked list. Figure 13 shows the
link that must be modified. However, there is a special case that needs
to be addressed. If the item to be removed happens to be the first item
in the list, then
current will reference the first node in the
linked list. This also means that
previous will be
None. We said
previous would be referring to the node whose next
reference needs to be modified in order to complete the remove. In this
case, it is not
previous but rather the head of the list that needs
to be changed (see Figure 14).
Line 12 allows us to check whether we are dealing with the special case
described above. If
previous did not move, it will still have the
None when the boolean
True. In that case
(line 13) the head of the list is modified to refer to the node after
the current node, in effect removing the first node from the linked
list. However, if previous is not
None, the node to be removed is
somewhere down the linked list structure. In this case the previous
reference is providing us with the node whose next reference must be
changed. Line 15 uses the
setNext method from
accomplish the removal. Note that in both cases the destination of the
reference change is
current.getNext(). One question that often
arises is whether the two cases shown here will also handle the
situation where the item to be removed is in the last node of the linked
list. We leave that for you to consider.
You can try out the
UnorderedList class in ActiveCode 1.
The remaining methods
left as exercises. Remember that each of these must take into account
whether the change is taking place at the head of the list or someplace
pop require that we name the
positions of the list. We will assume that position names are integers
starting with 0.
Part I: Implement the append method for UnorderedList. What is the time complexity of the method you created?
Part II: In the previous problem, you most likely created an append method that was \(O(n)\) If you add an instance variable to the UnorderedList class you can create an append method that is \(O(1)\). Modify your append method to be \(O(1)\) Be Careful! To really do this correctly you will need to consider a couple of special cases that may require you to make a modification to the add method as well.