We Keep Coding

Reversed double linked list by python

why can't print reversed this double linked list by python?
always print 6 or None
please can anyone help me fast to pass this task
///////////////////////////////////////////////////////////////////////////
class Node:
def __init__(self, data=None, next=None, prev=None):
self.data = data
self.next = next
self.previous = prev
sample methods==>
def set_data(self, newData):
self.data = newData
def get_data(self):
return self.data
def set_next(self, newNext):
self.next = newNext
def get_next(self):
return self.next
def hasNext(self):
return self.next is not None
def set_previous(self, newprev):
self.previous = newprev
def get_previous(self):
return self.previous
def hasPrevious(self):
return self.previous is not None
class double===>
class DoubleLinkedList:
def __init__(self):
self.head = None
self.tail = None
def addAtStart(self, item):
newNode = Node(item)
if self.head is None:
self.head = self.tail = newNode
else:
newNode.set_next(self.head)
newNode.set_previous(None)
self.head.set_previous(newNode)
self.head = newNode
def size(self):
current = self.head
count = 0
while current is not None:
count += 1
current = current.get_next()
return count
here is the wrong method ==>
try to fix it without more changes
def printReverse(self):
current = self.head
while current:
temp = current.next
current.next = current.previous
current.previous = temp
current = current.previous
temp = self.head
self.head = self.tail
self.tail = temp
print("Nodes of doubly linked list reversed: ")
while current is not None:
print(current.data),
current = current.get_next()
call methods==>
new = DoubleLinkedList()
new.addAtStart(1)
new.addAtStart(2)
new.addAtStart(3)
new.printReverse()

Your printReverse seems to do something else than what its name suggests. I would think that this function would just iterate the list nodes in reversed order and print the values, but it actually reverses the list, and doesn't print the result because of a bug.
The error in your code is that the final loop has a condition that is guaranteed to be false. current is always None when it reaches that loop, so nothing gets printed there. This is easily fixed by initialising current just before the loop with:
current = self.head
That fixes your issue, but it is not nice to have a function that both reverses the list, and prints it. It is better practice to separate these two tasks. The method that reverses the list could be named reverse. Then add another method that allows iteration of the values in the list. This is done by defining __iter__. The caller can then easily print the list with that iterator.
Here is how that looks:
def reverse(self):
current = self.head
while current:
current.previous, current.next = current.next, current.previous
current = current.previous
self.head, self.tail = self.tail, self.head
def __iter__(self):
node = self.head
while node:
yield node.data
node = node.next
def __repr__(self):
return "->".join(map(repr, self))
The main program can then be:
lst = DoubleLinkedList()
lst.addAtStart(1)
lst.addAtStart(2)
lst.addAtStart(3)
print(lst)
lst.reverse()
print(lst)

calling a method in another method errors

class Node:
def __init__(self, data):
self.data = data
self.ref = None
class LinkedList:
def __init__(self):
self.head = None
def show(self):
if self.head is None:
print("This linked lists is empty")
else:
currentnode = self.head
while currentnode is not None:
print(currentnode.data, end=" --> ")
currentnode = currentnode.ref
def addelement(self, value):
newnode = Node(value)
newnode.ref = self.head
self.head = newnode
def lenofll(self , i = 0):
while self.head is not None:
i = i +1
self.head = self.head.ref
return i
def middle(self):
i = 0
lent = self.lenofll()
if self.head is None: # self.head changed to None after calling lenofll method.
print("linked list is empty")
I wanted to get the length of linked lists in the middle method. But as I called self.lenofll(), it changed the self.head to None.
What can I do to fix this?

Indeed, doing self.head = self.head.ref modifies the head. You should not make any modifications to self.head in a method whose job is just to search in the list -- without modifying anything to it.
As you can see, that method keeps looping until self.head is not None is not true, i.e. when self.head is None. No wonder that self.head is None after running this method!
Use a local variable for this iteration instead:
def lenofll(self, i = 0):
node = self.head # use local variable
while node is not None:
i += 1
node = node.ref
return i

I added 4 elements in the linked list, but why only three are displayed and the last element is not printed

Here is my code. Please help me in finding out why the last element of the linkedlist is not getting printed and how can I print the last element too.
class Node(object):
def __init__(self, value):
self.value = value
self.next = None
class LinkedList(object):
def __init__(self, head = None):
self.head = head
def append(self, data):
current = self.head
if self.head:
while current.next:
current= current.next
current.next= data
else:
self.head = data
def show(self):
current = self.head
while current.next:
print(current.value)
current = current.next
ll1=LinkedList()
e1= Node(1)
e2 = Node(2)
e3 = Node(3)
e4 = Node(4)
ll1.append(e1)
ll1.append(e2)
ll1.append(e3)
ll1.append(e4)
ll1.show()
Image of output

Because e3.next is e4, but e4.next is None. So the while loop will end at e3.
Just change the show function like this:
def show(self):
current = self.head
while current.next:
print(current.value)
current = current.next
if not current.next:
print(current.value)

last element hasn't next element so not printed; you can change code to do-while format like bottom
def show(self):
current = self.head
while True:
print(current.value)
current = current.next
if not current:
break

Find the frequency of numbers using linkedlist

Find the frequency of numbers using linkedlist.
Getting SIGTSTP - time limit exceed error while running the below code. Can anyone help me where am I getting it wrong?
class Element(object):
def __init__(self,value):
self.value = value
self.next = None
class LinkedList(object):
def __init__(self, head = None):
self.head = head
def append(self, new):
current = self.head
if self.head:
while current.next:
current = current.next
current.next = new
else:
self.head = new
def traverse(self):
current = self.head
while current != None:
print(current.value)
current = current.next
arr = list(map(int, input().split()))
ll = LinkedList()
for i in arr:
e = Element(i)
ll.append(e)
ll.traverse()
def frequency(a):
current = a.head
while current != None:
count = 1
while current.next != None:
if current.value == current.next.value:
current+=1
if current.next.next != None:
current.next = current.next.next
else:
current.next = None
print(str(current.value)+" : " + str(count))
current = current.next
frequency(ll)

Everything looks fine except frequency. You will need to keep two references, one to the current element and the other will traverse the remainder of the list, starting from current. Does that give you something to go on?
Note too that your current implementation will modify the underlying linked list, while you can indeed do "skipping" with the pointers to prevent listing the same element multiple times, it is imo preferable to avoid modifying the underlying structure in this way.

constructing a Queue linked list using only one head pointer

I'm trying to construct a Queue linked list using only a head pointer (no tail).
but i cant seem to enqueue at the end of the list.
example: at the moment the code will: c -> b -> a, however i would like reverse it a -> b -> c.
class Node:
'''A node for a linked list.'''
def __init__(self, initdata):
self.data = initdata
self.next = None
class Queue(object):
def __init__(self):
self.head = None
def enqueue(self, item):
"""Add an item onto the tail of the queue."""
if self.head == None:
temp = Node(item)
temp.next = self.head
self.head = temp
else:
current = self.head
while current != None:
current = current.next
if current == None:
temp = Node(item)
temp.next = current
current = temp
def dequeue(self):
if self.head == None:
raise IndexError("Can't dequeue from empty queue.")
else:
current_first = self.head
current = self.head.next
return current_first.data

This should do it:
class Node:
'''A node for a linked list.'''
def __init__(self, initdata):
self.data = initdata
self.next = None
class Queue(object):
def __init__(self):
self.head = None
def enqueue(self, item):
"""Add an item onto the tail of the queue."""
if self.head is None:
self.head = Node(item)
else:
current = self.head
while current.next is not None:
current = current.next
current.next = Node(item)
def dequeue(self):
if self.head is None:
raise IndexError("Can't dequeue from empty queue.")
else:
first = self.head
self.head = self.head.next
return first.data
Besides some logic fixes (we need to create a new node and store it in current.next, current is just a variable pointing to a node), note we use is operator for testing for None and Node constructor to set data (so we can create and assign new nodes without temp var).
For example:
q = Queue()
q.enqueue('a')
q.enqueue('b')
q.enqueue('c')
print(q.dequeue())
print(q.dequeue())
print(q.dequeue())
Outputs:
a
b
c
Btw, note that such structure requires O(N) insertion time and O(1) deletion (pop) time. Double-ended queue (like the standard collections.deque) will do both insertion and deletion in constant time.