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)
Related
I'm trying to learn how to create linked lists. This is my first time doing this and the reason of code failure may be something basic I'm missing.
That being said, I am unable to figure out even after using vs code's debugger. It simply stops at the end of the append method when it is called the second time.
I am using recursion to traverse to the tail. Could that be a the problem?
class Node:
def __init__(self, data, next_node=None):
self.data = data
self.next = next_node
class LinkedList:
def __init__(self):
self.head = None
def __repr__(self):
if not self.head:
return 'Linked list is empty'
linked_list = self.head.data
if self.head.next == None:
return linked_list
current = self.head
while current.next != None:
linked_list += '\n|\nV' + current.data
return linked_list
def append(self, value):
if not self.head:
self.head = Node(data=value)
return
tail = self.tail()
tail.next = Node(data=value)
def tail(self):
tail = self._traverse_to_tail(self.head)
while tail.next != None:
tail = self._traverse_to_tail(tail)
return tail
def _traverse_to_tail(self, current_node, recursion_count=0):
print(current_node.data)
if recursion_count > 997:
return current_node
if current_node.next == None:
return current_node
current_node = current_node.next
recursion_count += 1
return self._traverse_to_tail(current_node, recursion_count)
if __name__ == '__main__':
ll = LinkedList()
ll.append('foo')
ll.append('baz')
print(ll)
The problem is you have an infinite loop in the __repr__() function, because you never increment current.
def __repr__(self):
if not self.head:
return 'Linked list is empty'
linked_list = self.head.data
if self.head.next == None:
return linked_list
current = self.head
while current.next != None:
current = current.next
linked_list += '\n|\nV' + current.data
return linked_list
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'm new to python coming form c++ i don't know how to work with linked list without pointers, that being said I've written this code but it returns the same list without sorting it at all
class ListNode:
def __init__(self, val=0, next=None):
self.val = val
self.next = next
def insertionSortList(head):
if head.next==None:
return head
#checkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkks
loop=head
i=head
while(i.next!=None):
save_val=i.next
i=i.next.next
while True:
if loop!=i and save_val.val>loop.val:
loop=loop.next
else:
if loop==head:
save_val=loop
break
else:
save_val=loop.next
loop=save_val
loop=head
break
return head
You are not incrementing the loop node variable. Your code breaks out of the inner while loop before the loop node variable gets a chance to increment loop=loop.next. Here is my solution, I separated the Node and Linked List classes. Setup a length variable and incremented the value each time I inserted a node. Then used the normal insertion sort method.
class Node:
def __init__(self,value):
self.value = value
self.next = None
class ListNode:
def __init__(self):
self.head = None
self.length = 0
def __repr__(self):
'''Allows user to print values in the ListNode'''
values = []
current = self.head
while current:
values.append(current.value)
current = current.next
return ', '.join(str(value) for value in values)
def insert(self,value):
'''Inserts nodes into the ListNode class'''
newNode = Node(value)
if self.head is None:
self.head = newNode
else:
newNode.next = self.head
self.head = newNode
self.length += 1
return self
def insertionSortList(self):
'''Insertion sort method: held a temp variable and inserted the smallest value from temp through the rest of the list then incremented temp variable to the next idx/node'''
current = self.head
for _ in range(self.length-1):
tempNode = current.next
while tempNode:
if tempNode.value < current.value:
current.value, tempNode.value = tempNode.value, current.value
tempNode = tempNode.next
current = current.next
return self
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.
I am trying to implement an algorithm to remove duplicates from a linkedlist but my algorithm freeze when it comes to check if the current node has data equal with the next one.
class Node:
def __init__(self, data, next):
self.data = data
self.next = next
class LinkedList:
def __init__(self, head = None):
self.head = head
def add(self, item):
newNode = Node(item, self.head)
self.head = newNode
def printit(self):
current = self.head
while current is not None:
print(current.data)
current = current.next
def removeDuplicates(self):
current = self.head
while current != None:
runner = current
while runner.next != None:
if runner.next.data == current.data:
runner.next = current.next.next
else:
runner = current.next
current = current.next
mylist = LinkedList()
mylist.add(31)
mylist.add(77)
mylist.add(31)
mylist.add(22)
mylist.add(22)
mylist.add(22)
mylist.printit()
mylist.removeDuplicates()
mylist.printit()
It is maybe really silly, but I can't spot it right now, any ideas?
In your while loop:
runner = current.next
should be
runner = runner.next
otherwise it just keeps taking the same current.next on each iteration, since it never changes current in the loop. Same fix a couple of lines earlier.