Device driver: System software or application software
2 answers:
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Answer:
A python programming language (code) was used for this exercise.
The code is shown below in the explanation section
Explanation:
Solution
Python code:
class LN:
def __init__(self,value,next=None):
self.value = value
self.next = next
def list_to_ll(self,l):
if l == []:
return None
front = rear = LN(l[0])
for v in l[1:]:
rear.next = LN(v)
rear = rear.next
return front
def str_ll(self,ll):
answer = ''
while ll != None:
answer += str(ll.value)+'->'
ll = ll.next
return answer + 'None'
def append_ordered(self,ll,v):
ln1=LN(v); #create a node with the new value by creating an instance of the LN class
if ll is None: #if the linked list is empty
ll=ln1 #return the new created node
elif(ll.value >= ln1.value): #if the value to be add is smallest, append the linked list to it
ln1.next=ll
ll=ln1
else:
current=ll; #crate a temporary pointer to iterate over linked list
#iterate till the next node value is smaller than the new value and list is not over
while (current.next is not None and current.next.value < ln1.value):
current=current.next
ln1.next=current.next #save the current node in the next of the new node
current.next=ln1 #save new node in the next of current node
return ll
ln=LN(0); #create an instance of class LN
ll1=ln.list_to_ll([1,2,4,5]); #create a linked list of the list
answer=ln.str_ll(ll1);
print(answer); #print the linked list
ll2=ln.append_ordered(ll1,3); #append the value 3 in the linked list
answer=ln.str_ll(ll2)
print(answer); #print the new linked list
Now,
- If Linked list is empty then make the node as head and return it.
- If value of the node to be inserted is smaller than value of head node, then insert the node at start and make it head.
- In a loop, find the appropriate node after which the input node ( 10 ) is to be inserted.To find the appropriate node start from head,keep moving until you reach a node GN (12 ) who's value is greater than the input node. The node just before GN is the appropriate node (8).
- Insert the node (10) after the appropriate node (8) found in step 3.
Note: Kindly find the output python code attached to the question below.
