Introduction
So far, in this 3-part series about linked lists in Python, we started our discussion about the linked list. We saw what the linked list is along with its advantages and disadvantages. We also studied some of the most commonly used linked list methods such as traversal, insertion, deletion, searching, and counting an element. Finally, we saw how to reverse a linked list.
In this article, we will continue from where we left in the last article and will see how to sort a linked list using bubble and merge sort, and how to merge two sorted linked lists.
Before we continue, it is imperative to mention that you should create the Node
and LinkedList
classes that we created in the last article.
Our 3-part series about linked lists in Python:
Sorting a Linked Lists using Bubble Sort
There are two ways to sort a linked list using bubble sort:
- Exchanging data between nodes
- Modifying the links between nodes
In this section, we will see how both these approaches work. We will use the bubble sort algorithm to first sort the linked list by changing the data, and then we will see how we can use bubble sort to change the links in order to sort the linked list.
Sorting Linked List by Exchanging Data
To sort a linked list by exchanging data, we need to declare three variables p
, q
, and end
. The variable p
will be initialized with the start node, while the end
will be set to None
.
Note: It is important to remember that to sort the list with n
elements using bubble sort, you need n-1
iterations.
To implement bubble sort, we need two while
loops. The outer while
loop executes until the value of the variable end
is equal to the self.start_node
.
The inner while
loop executes until p
becomes equal to the end
variable. Inside the outer while
loop, the value of p
will be set to self.start_node
which is the first node. Inside the inner while
loop, the value of q
will be set to p.link
which is actually the node next to q
. Then the values of p
and q
will be compared if p
is greater than q
the values of both the variables will be swapped and then p
will point to p.ref
, which is the next node. Finally, the end
will be assigned the value of p
. This process continues until the linked list is sorted.
Let's understand this process with the help of an example. Suppose we have the following list:
8,7,1,6,9
Let's implement our algorithm to sort the list. We'll see what will happen during each iteration.
The purpose of the bubble sort is that during each iteration, the largest value should be pushed to the end, hence at the end of all iterations, the list will automatically be sorted.
Before the loop executes, the value of end
is set to None
.
In the first iteration, p
will be set to 8
, and q
will be set to 7
. Since p
is greater than q
, the values will be swapped and p
will become p.ref
. At this point in time, the linked list will look like this:
7,8,1,6,9
From now on, p
is not equal to the end
, the loop will continue and now p
will become 8
and q
will become 1
. Since p
is again greater than q
, the values will be swapped again and p
will again become p.ref
:
7,1,8,6,9
Here again, p
is not equal to the end
, the loop will continue and now p
will become 8
and q
will become 6
. Since again p
is greater than q
, the values will be swapped again and p
will again become p.ref
. The list will look like this:
7,1,6,8,9
Again p
is not equal to the end
, the loop will continue and now p
will become 8
and q
will become 9
. Here since p
is not greater than q
, the values will not be swapped and p
will become p.ref
. At this point in time, the reference of p
will point to None
, and the end
also points to None
. Hence the inner while
loop will break and the end
will be set to p
.
In the next set of iterations, the loop will execute until 8
, since 9
is already at the end. The process continues until the list is completely sorted.
The Python code for sorting the linked list using bubble sort by exchanging the data is as follows:
def bub_sort_datachange(self):
end = None
while end != self.start_node:
p = self.start_node
while p.ref != end:
q = p.ref
if p.item > q.item:
p.item, q.item = q.item, p.item
p = p.ref
end = p
Add the bub_sort_dataexchange()
method to the LinkedList
class that you created in the last article.
Once you add the method to the linked list, create any set of nodes using the make_new_list()
method and then use the bub_sort_dataexchange()
to sort the list. You should see the sorted list when you execute the traverse_list()
method.
Sorting Linked Lists by Modifying Links
Bubble sort can also be used to sort a linked list by modifying the links instead of changing data. The process remains quite similar to sorting the list by exchanging data, however, in this case, we have an additional variable r
that will always correspond to the node previous than the p
node.
Let's take a simple example of how we will swap two nodes by modifying links. Suppose we have a linked list with the following items:
10,45,65,35,1
And we want to swap 65
and 35
. At this point in time p
corresponds to node 65
, and q
corresponds to node 35
. The variable r
will correspond to node 45
(previous to node p
). Now if the node p
is greater than node q
, which is the case here, the p.ref
will be set to q.ref
and q.ref
will be set to p
. Similarly, r.ref
will be set to q
. This will swap nodes 65
and 35
.
The following method implements the bubble sorting for the linked list by modifying links:
def bub_sort_linkchange(self):
end = None
while end != self.start_node:
r = p = self.start_node
while p.ref != end:
q = p.ref
if p.item > q.item:
p.ref = q.ref
q.ref = p
if p != self.start_node:
r.ref = q
else:
self.start_node = q
p,q = q,p
r = p
p = p.ref
end = p
Add the bub_sort_linkchange()
method to the LinkedList
class that you created in the last article.
Once you add the method to the linked list, create any set of nodes using the make_new_list()
method and then use the bub_sort_linkchange()
to sort the list. You should see the sorted list when you execute the traverse_list()
method.
Merging Sorted Linked List
In this section, we will see how we can merge two sorted linked lists in a manner that the resulting linked list is also sorted. There are two approaches to achieve this. We can create a new linked list that contains individually sorted lists or we can simply change the links of the two linked lists to join the the two. In the second case, we do not have to create a new linked list.
Let's first see how we can merge two linked lists by creating a new list.
Merging Sorted Linked Lists by Creating a New List
Let's first dry-run the algorithm to see how we can merge two sorted linked lists with the help of a new list.
Suppose we have the following two sorted linked lists:
list1:
10,45,65
list2:
5,15,35,68
These are the two lists we want to merge. The algorithm is straightforward. All we will need is three variables, p
, q
, and em
, and an empty list newlist
.
At the beginning of the algorithm, p
will point to the first element of list1
whereas q
will point to the first element of the list2
. The variable em
will be empty. At the start of the algorithm, we will have the following values:
p = 10
q = 5
em = None
newlist = None
Next, we will compare the first element of the list1
with the first element of list2
, in other words, we will compare the values of p
and q
and the smaller value will be stored in the variable em
which will become the first node of the new list. The value of em
will be added to the end of the newlist
.
After the first comparison, we will have the following values:
p = 10
q = 15
em = 5
newlist = 5
Since q
was less than p
, we stored the value of q
in em
and moved q
one index to the right. In the second pass, we will have the following values:
p = 45
q = 15
em = 10
newlist = 5, 10
Here since p
was smaller, we added the value of p
to newlist
, set em
to p
, and then moved p
one index to the right:
p = 45
q = 35
em = 15
newlist = 5, 10, 15
Similarly, in the next iteration:
p = 45
q = 68
em = 35
newlist = 5, 10, 15, 35
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In the next iteration, p
will again be smaller than q
, hence:
p = 65
q = 68
em = 45
newlist = 5, 10, 15, 35, 45
And, finally:
p = None
q = 68
em = 65
newlist = 5, 10, 15, 35, 45, 65
When one of the lists becomes None
, all the elements of the second list are added at the end of the new list. Therefore, the final list will be:
p = None
q = None
em = 68
newlist = 5, 10, 15, 35, 45, 65, 68
Let's put all this into code by creating two methods - merge_helper()
and merge_by_newlist()
:
def merge_helper(self, list2):
merged_list = LinkedList()
merged_list.start_node = self.merge_by_newlist(self.start_node, list2.start_node)
return merged_list
def merge_by_newlist(self, p, q):
if p.item <= q.item:
startNode = Node(p.item)
p = p.ref
else:
startNode = Node(q.item)
q = q.ref
em = startNode
while p is not None and q is not None:
if p.item <= q.item:
em.ref = Node(p.item)
p = p.ref
else:
em.ref = Node(q.item)
q = q.ref
em = em.ref
while p is not None:
em.ref = Node(p.item)
p = p.ref
em = em.ref
while q is not None:
em.ref = Node(q.item)
q = q.ref
em = em.ref
return startNode
The merge_helper()
method takes a linked list as a parameter and then passes the self
class, which is a linked list itself, and the linked list passed to it as a parameter, to the merge_by_newlist()
method.
The merge_by_newlist()
method merges the two linked lists by creating a new linked list and returns its start node. Add these two methods to the LinkedList
class. Create two new linked lists, sort them using the bub_sort_datachange()
or the bub_sort_linkchange()
methods that you created in the last section, and then use the merge_by_newlist()
to see if you can merge two sorted linked lists or not.
Merging Sorted Linked Lists by Rearranging Links
In this approach, a new linked list is not used to store the merger of two sorted linked lists. Rather, the links of the two linked lists are modified in such a way that the two linked lists are merged in a sorted manner.
Let's see a simple example of how we can do this. Suppose we have the same two lists list1
and list2
:
list1:
10,45,65
list2:
5,15,35,68
We want to merge them in a sorted manner by rearranging the links. To do so we need variables p
, q
, and em
. Initially, they will have the following values:
p = 10
q = 5
em = none
newlist = none
Next, we will compare the first element of list1
with the first element of list2
, in other words, we will compare the values of p
and q
and the smaller value will be stored in the variable em
which will become the first node of the new list.
After the first comparison, we will have the following values:
p = 10
q = 15
start = 5
em = start
After the first iteration, since q
is less than p
, the start node will point towards q
and q
will become q.ref
. The em
will be equal to the start
. The em
will always refer to the newly inserted node in the merged list:
p = 45
q = 15
em = 10
Here, since p
was smaller than the q
, the variable em
now points towards the original value of p
and p
becomes p.ref
:
p = 45
q = 35
em = 15
Since q
was smaller than p
, em
points towards q
and q
becomes q.ref
:
p = 45
q = 68
em = 35
Similarly em
here points towards q
:
p = 65
q = 68
em = 45
newlist = 5, 10, 15, 35, 45
And here em
points towards becomes p
:
p = None
q = 68
em = 65
newlist = 5, 10, 15, 35, 45, 65
When one of the lists becomes None
, the elements from the second list are simply added at the end:
p = None
q = None
em = 68
newlist = 5, 10, 15, 35, 45, 65, 68
The script that contains methods for merging two lists without creating a new list is as follows:
def merge_helper2(self, list2):
merged_list = LinkedList()
merged_list.start_node = self.merge_by_linkChange(self.start_node, list2.start_node)
return merged_list
def merge_by_linkChange(self, p, q):
if p.item <= q.item:
startNode = Node(p.item)
p = p.ref
else:
startNode = Node(q.item)
q = q.ref
em = startNode
while p is not None and q is not None:
if p.item <= q.item:
em.ref = Node(p.item)
em = em.ref
p = p.ref
else:
em.ref = Node(q.item)
em = em.ref
q = q.ref
if p is None:
em.ref = q
else:
em.ref = p
return startNode
In the script above we have two methods: merge_helper2()
and merge_by_linkChange()
. The first method merge_helper2()
takes a linked list as a parameter and then passes the self class which is a linked list itself and the linked list passed to it as a parameter, to the merge_by_linkChange()
, which merges the two linked by modifying the links and returns the start node of the merged list.
Add these two methods to the LinkedList
class. Create two new linked lists, sort them using the bub_sort_datachange()
or the bub_sort_linkchange()
methods that you created in the last section, and then use the merge_by_newlist()
to see if you can merge two sorted linked lists or not. Let's see this process in action:
new_linked_list1 = LinkedList()
new_linked_list1.make_new_list()
The script will ask you for the number of nodes to enter. Enter as many nodes as you like and then add values for each node as shown below:
How many nodes do you want to create: 4
Enter the value for the node:12
Enter the value for the node:45
Enter the value for the node:32
Enter the value for the node:61
Next, create another linked list repeating the above process:
new_linked_list2 = LinkedList()
new_linked_list2.make_new_list()
Next, add a few dummy nodes with the help of the following script:
How many nodes do you want to create: 4
Enter the value for the node:36
Enter the value for the node:41
Enter the value for the node:25
Enter the value for the node:9
The next step is to sort both the lists. Execute the following script:
new_linked_list1. bub_sort_datachange()
new_linked_list2. bub_sort_datachange()
Finally, the following script merges the two linked lists:
list3 = new_linked_list1.merge_helper2(new_linked_list2)
To see if the lists have actually been merged, execute the following script:
list3.traverse_list()
The output looks like this:
9
12
25
32
36
41
45
61
Conclusion
In this article, we continued from where we left in the previous article. We saw how we can sort merge lists by changing data and then modifying links. Finally, we also studied different ways of merging two sorted linked lists.
In the next article, we'll take a look at how to construct and perform operations on doubly linked lists.