 Sorting and Merging Single Linked List - Stack Abuse

# Sorting and Merging Single Linked List In the last article, 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 method 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 Node and LinkedList classes that we created in the last article.

### Sorting a Linked List using Bubble Sort

There are two ways to sort a linked list using bubble sort:

1. Exchanging data between nodes
2. 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 end will be set to None.

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 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 of time the linked list will look like this:

7,8,1,6,9


Since at this point of time, p is not equal to end, the loop will continue and now p will become 8 and q will become 1. 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:

7,1,8,6,9


Here again, p is not equal to 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 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 of time, the reference of p will point to None, and end also points to None. Hence the inner while loop will break and 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() function and then use the bub_sort_dataexchange() to sort the list. You should see the sorted list when you execute the traverse_list() function.

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() function and then use the bub_sort_linkchange() to sort the list. You should see the sorted list when you execute the traverse_list() function.

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 links of the two linked list to join the two sorted linked list. 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 list 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 straight forward. 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 the 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, therefore, we store the value of q in em 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 add the value of p to newlist, and set em to p and then moved p one index to the right. In the next iteration we have:

p = 45
q = 35
em = 15
newlist = 5, 10, 15


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Similarly, in the next iteration:

p = 45
q = 68
em = 35
newlist = 5, 10, 15, 35


And in the next iteration, p will again be smaller than q, hence:

p = 65
q = 68
em = 45
newlist = 5, 10, 15, 35, 45


Finally,

p = None
q = 68
em = 65
newlist = 5, 10, 15, 35, 45, 65


When one of the list 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


The Python script for merging two sorted lists is as follows:

    def merge_helper(self, list2):
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


In the script above we have two methods: merge_helper() and merge_by_newlist(). The first method merge_helper() 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 by creating a new linked list and returns the start node of the new linked 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.

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 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 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.

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 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


Here 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 functions for merging two lists without creating a new list is as follows:

    def merge_helper2(self, list2):
return merged_list

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.

Create a new linked list using the following script:

new_linked_list1 = LinkedList()


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()


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()


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 my 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.

Last Updated: February 28th, 2019

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Usman MalikAuthor

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