Hey everyone I am trying to print (2->None, 3) in the linked list below, but am getting a syntax error. can someone please help me with this syntax cannot find on google for some reason. Code below:
class Node(object):
def __init__(self, data=None, next_node=None):
self.data = data
self.next = next_node
def Insert(head, data):
if head == None:
head = Node(data)
print(head)
else:
current = head
while current.next != None:
current = current.next
current.next = Node(data)
return head
print Insert(2->None, 3) # -> is bringing a syntax error, how do I write this in python 2.7?
Honestly, I think your confused about how to implement a linked list in the first place. If you want a linked list in the form of:
[2] -> [3] -> [None]
Then you need to insert each element backwards. First None then 3 then 2. You also need to make your insert method inside of a class because you need to save state. Here is what I'd recommend instead:
class Node(object):
def __init__(self, data, next_node=None):
self.data = data
self.next_node = next_node
# Create a class not a function because we need to save state
# More specficly, we need to create a "global" variable which
# keeps track of the head of the linked list.
class LinkedList(object):
def __init__(self):
self.head = None
# Put the insert function inside of of the class.
# That way, we can save and load the state of self.head.
def insert(self, data):
new_node = Node(data, self.head)
self.head = new_node
# demo
ll = LinkedList()
# insert the elements in the reverse order you want
# them to appear.
ll.insert(None)
ll.insert(3)
ll.insert(2)
print("Head:", ll.head.data) # Head: 2
print("Middle:", ll.head.next_node.data) # Middle: 3
print("Tail:", ll.head.next_node.next_node.data) # Tail: None
Also, I recommend to a little research as well. Such as this article. Or just google Linked List in Python and browse over some of the results.
I have a feeling you just want to insert None as the first argument as what #leaf said. In that case, just directly use None:
print Insert(None, 3)
-> doesn't exist in Python as of now. This is #leaf's comment:
If you want to pass in none, simply pass in None: Insert(None, 3). Python doesn't have pointer in the sense your thinking about.
Related
I tried program to delete node from linked list recursively. My program is given below
class node:
def __init__(self, data=None):
self.data = data
self.next = None
class linkedList:
def __init__(self):
self.head=None
def printList(self):
cur = self.head
while(cur != None):
print(cur.data,end="->")
cur=cur.next
print("null")
def push(self,dat):
newNode = node(dat)
temp = self.head
self.head = newNode
newNode.next = temp
#staticmethod
def dnr(head, key):
if head is not None:
if head.data == key:
head=head.next
return head
if head is not None:
head.next = linkedList.dnr(head.next, key)
return head
return head
if __name__ == '__main__':
ll=linkedList()
ll.push(3)
ll.push(6)
ll.push(9)
ll.push(12)
ll.push(15)
ll.printList()
print("*******")
# ll.head = linkedList.dnr(ll.head,2)
linkedList.dnr(ll.head,9)
ll.printList()
The problem with this is that this does not work for first element.To make it work for first element I have to call the function like this
ll.head = linkedList.dnr(ll.head,2)
second thing is that I wanted my function to call this way
ll.dnr(2)
please tell me how to create a recursive function to delete node in linked list in python
I rewrote your code:
class node:
def __init__(self, data=None):
self.data = data
self.next = None
class linkedList:
def __init__(self):
self.__head=None
def printList(self):
cur = self.__head
while(cur != None):
print(cur.data,end="->")
cur=cur.next
print("null")
def push(self,dat):
newNode = node(dat)
temp = self.__head
self.__head = newNode
newNode.next = temp
#staticmethod
def __dnr(head, key):
if head is None:
return head
if head.data == key:
head = head.next
return head
head.next = linkedList.__dnr(head.next, key)
return head
#staticmethod
def dnr(listObj, key):
if listObj is None or listObj.__head is None:
return listObj
if listObj.__head.data == key:
listObj.__head = listObj.__head
listObj.__head = linkedList.__dnr(listObj.__head, key)
def deleteKey(self, key):
linkedList.dnr(self, key)
if __name__ == '__main__':
ll=linkedList()
ll.push(3)
ll.push(6)
ll.push(9)
ll.push(12)
ll.push(15)
ll.printList()
print("*******")
linkedList.dnr(ll, 9)
ll.deleteKey(12)
ll.printList()
I made head variable inside linkedList class private, it's not smart to give access outer world to class core components. And class core components should never leak to the outer world because if it's that not used properly that can cause errors. So I rewrote your dnr function and now it's more clear and it doesn't return head object which is core component of linkedList class. Now dnr function just check if passed listObj is valid and check if head is that node that should be deleted. After that it calls private static __dnr function to delete node with given key. Function deleteKey can be called like this ll.deleteKey(12), that is what you wanted.
Giving core component accessible through outer world is like giving bank customer access to a bank vault. Not everybody will try to steal money from it, but there will be someone who will try.
If you don't understand private variables follow this link.
I wanted my function to call this way ll.dnr(2)
Then you need to define an instance method. Your static function can serve a purpose, but as you noted, you really need to assign its return value back to your list's head attribute to be sure it also works when the original head node is removed. With your static method you cannot avoid this overhead, since that method has no knowledge about your linkedList instance.
You can achieve what you want simply by adding an instance method, that will rely on the existing static method and will deal with this assignment back to the instance's head attribute.
Add this to your linkedList class:
def remove(self, key):
self.head = linkedList.dnr(self.head, key)
Now in your main program you can do:
ll.remove(15)
Side note: you don't need the second if head is not None: check in your static method, as this condition will always be true when the execution reaches that point in your code. Just do the assignment to head.next unconditionally.
Addendum
If you want dnr itself to become an instance method (without addition of a remove method), then you need to temporarily cut off the head node from the list (even if you want to keep it), recur, and then conditionally add that cut-off node again (if it's key is not the one to delete).
It would look like this:
def dnr(self, key):
head = self.head
if head:
self.head = head.next # Skip
if head.data != key: # Need to restore it
self.dnr(key) # But first look further...
head.next = self.head # Prefix the removed node
self.head = head # ...and make it the head again
You would call like:
ll.dnr(15)
creating a linked list
inserting new data in linked list at end
printing linked list in list
Expected output: [5,6,2,8]
class node():
def __init__(self, data):
self.data = data
self.next = None
class linkedlist():
def add_at_end(self,newval,head):
newnode = node(newval)
while head.next:
head = head.next
head.next = newval
def print_list(self,head):
LL = []
while head:
LL.append(head.data)
head = head.next
return LL
x = node(5)
y = node(6)
z = node(2)
x.next = y
y.next = z
vals = linkedlist()
print(vals.add_at_end(8,x))
print(vals.print_list(x))
The error occurs because you have head.next = newval where newval is a number, but you should be assigning a node instance, i.e. newnode. So the next time you iterate the list, like in print_list you'll bump into that integer, and access its data attribute, as if it is a node...
Some other remarks on your code:
Your linkedlist class has no state. It merely provides some functions that act on the head argument. The OOP way to do this, is to give a linkedlist instance its own head attribute. And then rely on the self argument, which in your current code is not used in the functions.
Your main code should not have to deal with node instances. It should just have to create a linkedlist, and then populate it by calling its method(s). For that to work you'll have to deal with the case where the list is empty, and a first value is added to it, because then its head attribute needs to be set to it.
Use PascalCase for naming your classes.
The name of the print_list function is misleading: it doesn't print, but returns a list. So I would name it to_list instead.
Don't print the result of calling vals.add_at_end, because it is not designed to return anything, so you'll be printing None.
Corrected:
class Node(): # Use PascalCase
def __init__(self, data):
self.data = data
self.next = None
class LinkedList():
def __init__(self):
self.head = None # Give each instance its own head attribute
def add_at_end(self, newval):
if not self.head: # Deal with this case
self.head = Node(newval)
else:
head = self.head # Use attribute
while head.next:
head = head.next
head.next = Node(newval) # corrected
def to_list(self): # Better name
LL = []
head = self.head # Use attribute
while head:
LL.append(head.data)
head = head.next
return LL
vals = LinkedList()
vals.add_at_end(5) # Use method to add nodes
vals.add_at_end(6)
vals.add_at_end(2)
vals.add_at_end(8) # Nothing to print here
print(vals.to_list())
I'm writing a linked list in Python and I've come across an issue which is really troublesome and terrible for debugging and I feel like I'm missing something about Python. I'm supposed to create a singly linked list with some basic functionalities. One of them is the take() function, which is meant to create a new list of n first elements of the original list.
However, creating a new instance of the LinkedList class seems to change the .self parameter and the variable node is modified, as the attribute .next is turned to None. In result, when creating a list and then trying to make a new one out of the n elements of it, the program runs indefinitely, but no matter which part I look at, I cannot find the loop or the reason behind it.
class LinkedList:
def __init__(self, head=None):
self.head = head
def is_empty(self):
if self.head == None:
return True
else:
return False
def add_last(self, node):
if self.is_empty():
self.head = node
return
nextEl = self.head
while True:
if nextEl.next is None:
nextEl.next = node
return
nextEl = nextEl.next
def take(self, n):
node = self.head
newHead = self.head
newHead.next = None
newList = LinkedList(newHead)
count = 0
while count < n:
newList.add_last(node.next)
node = node.next
count += 1
return newList
class Node:
def __init__(self, data, next=None):
self.data = data
self.next = next
Thank you for all your help.
In the take() function the line
newHead.next = None
modifies a node of the linked list, breaking this list. You can fix this as follows:
def take(self, n):
node = self.head
newHead = Node(self.head.data)
newList = LinkedList(newHead)
count = 0
while count < n:
newList.add_last(node.next)
node = node.next
count += 1
return newList
This, however, still will not work correctly since there is a problem with the add_last() function too. This function, I think, is supposed to add a node as the last element of a linked list, but since you do not modify the next attribute of the node, you actually append a whole linked list starting with that node. This can be fixed in the following way:
def add_last(self, node):
if self.is_empty():
self.head = node
return
nextEl = self.head
while True:
if nextEl.next is None:
nextEl.next = Node(node.data)
return
nextEl = nextEl.next
There are more issues. For example, take(sefl, n) will actually create a list of n+1 elements and will throw an exception if it is applied to a linked list that does not have that many elements.
I am brushing up on some data structures and algorithms with Python, so I am implementing an Unordered Linked list. Within the same file I first wrote a Node class followed by a List class. What I don't get is how the "current" variable in my search_item() method seems to be a node object or at least able to access the Node class methods and attributes. I noticed that if I comment out my add_node() method then "current" no longer has access to Node's methods. Now I am not explicitly using neither inheritance nor composition, so I am having a hard time seeing how current just gets to call get_next() the way the code is written below. I would think I'd have to declare current as: current = Node(self.head) but just current = self.head seems to work?
Your help would be greatly appreciated.
class Node:
def __init__(self, data):
self.data = data
self.next = None
def get_data(self):
return self.data
def set_data(self, d):
self.data = d
def get_next(self):
return self .next
def set_next(self, n):
self.next = n
class UnorderedList:
def __init__(self):
self.head = None
def add_node(self, item):
tmp = Node(item)
tmp.set_next(self.head)
self.head = tmp
def search_item(self, item):
current = self.head
# current = Node(self.head)
found = False
while current != None and not found:
if current.get_data() == item:
found = True
else:
current = current.get_next()
return found
Well if you comment out add_node then you do not add nodes to your linked list any longer therefore search_item will always see the initial value of self.head which is None.
Calling current.get_next() just works because via add_node you always ensure that self.head points either to None or to an instance of Node as tmp is created by tmp = Node(item) and then assigned to self.head = tmp. Therefore when setting current = self.head it will already refer to an instance of Node (or None) so you do not need to call current = Node(self.head).
I just recently came across the concept of duck typing and this old post of mine came to mind. It seems that this is what's at play and just didn't understand it at the time. 'current' is set to None by default, by when invoking any of the methods defined in the Node class, it is automatically defined as Node object.
There's this HackerRank problem about implementing a linked list. It is pretty basic and I thought I'd do it in a jiffy since I'd done all kinds of linked list implementations in C++. But I'm stuck somewhere.
class Node:
def __init__(self,data):
self.data = data
self.next = None
class Solution:
def display(self,head):
current = head
while current:
print current.data,
current = current.next
def insert(self,head,data):
new_node = Node(data)
if head == None:
head = new_node
else:
current = head
while current.next:
current = current.next
current.next = new_node
mylist= Solution()
T=int(input())
head=None
for i in range(T):
data=int(input())
head=mylist.insert(head,data)
mylist.display(head)
Only the insert function is editable. The rest of the code is provided by HackerRank and cannot be changed. The code doesn't print anything at the end of the insertions, and when I tried to print out the values while inserting, it seemed like the head kept on moving forwards instead of staying at the beginning.
You are appending the new_node after all the elements, and also you are not returning the newly created node. Modify your insert method like this:
def insert(self,head,data):
new_node = Node(data)
if head is not None:
current = head
new_node.next = current
return new_node
I think the error is here:
head = mylist.insert(head, data)
the method Solution.insert() is not returning anything, therefore head is assigned None every time.
When done, Solution.insert() must return the new_node created
The reason your code doesn't work is because of the None check that you are doing in insert method.
if head == None:
head = new_node
The head you are updating is local to the function insert and doesn't change head in the loop. So, head is always None in your loop.