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
Related
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)
I'm working on this problem: https://leetcode.com/explore/learn/card/linked-list/209/singly-linked-list/1290/
and leetcode is telling me that my code is throwing a runtime error. The input it's trying to make my code run is this:
["MyLinkedList","addAtHead","deleteAtIndex","addAtHead","addAtHead","addAtHead","addAtHead","addAtHead","addAtTail","get","deleteAtIndex","deleteAtIndex"]
[[],[2],[1],[2],[7],[3],[2],[5],[5],[5],[6],[4]]
Where the first array contains the function name and the second contains the function parameters. E.g. the second element adds a 2 at the head of the linkedlist. My question is that how would it be OK to be deleting the element at index 1 when there is only one element in the linkedlist? Is there some special thing about linked lists that I should know or is the input case incorrect. Below is my code for reference.
class SinglyLinkedList():
def __init__(self):
self.head = None
self.tail = None
self.length = 0
def get(self, index):
if index<0 or index>self.length: raise AssertionError("get index must be in bounds")
node = self.head
for _ in range(index):
node = node.next
return node
def addAtHead(self, val):
node = SinglyListNode(val)
if self.length==0:
self.head = node
self.tail = self.head
else:
node.next = self.head
self.head = node
self.length+=1
def addAtTail(self, val):
node = SinglyListNode(val)
if self.length==0:
self.tail = node
self.head = self.tail
else:
self.tail.next = node
self.tail = node
self.length+=1
def addAtIndex(self, index, val):
if index<0 or index>self.length:
raise AssertionError(f"index at which to add ({index}) is out of bounds")
if index==0:
return self.addAtHead(val)
if index==self.length:
return self.addAtTail(val)
newNode = SinglyListNode(val)
node = self.head
for _ in range(index-1):
node = node.next
newNode.next = node.next
node.next = newNode
self.length+=1
def deleteAtIndex(self, index):
if index<0 or index>self.length:
raise AssertionError(f"index at which to add ({index}) is out of bounds")
if index==0:
self.head = self.head.next
elif index==self.length-1:
self.tail=self.get(self.length-2)
else:
node = self.head
for _ in range(index-1):
node = node.next
node.next = node.next.next
self.length-=1
def __str__(self):
res = "["
node = self.head
for _ in range(self.length-1):
res += str(node)+","
node = node.next
res += str(node)+f"] ({self.length})"
return res
From https://leetcode.com/problems/design-linked-list/
void deleteAtIndex(int index) Delete the indexth node in the linked list, if the index is valid.
Looks like the delete needs to be done only when the index is valid. Your code needs to handle the case when the index is invalid.
I want to select the first node in the linked list and present the selected node. This is the whole code that I've created. The "prepend" adds the node before the first node. The "append" adds the node after the last of the linked list.
class Node:
def __init__(self, data=None, next=None):
self.data = data
self.next = next
def __str__(self):
return str(self.data)
# LinkedList definition here
class LinkedList:
def __init__(self):
self.head = None
self.tail = None
def prepend(self, data):
node = Node(data, self.head)
if self.head is None:
self.head = node
self.tail = node
else:
node.next = self.head
self.head = node
def append(self, data):
node = Node(data, None)
if self.tail is None:
# No elements in list
self.head = node
self.tail = node
else:
self.tail.next = node
self.tail = node
def pop_start(self):
if self.head is None:
return None
if self.head.next is None:
cur = self.head
self.head = None
return cur
else:
if self.head != None:
temp = self.head
self.head = self.head.next
return temp
names = LinkedList()
names.append("Bill Gates")
names.append("Steve Jobs")
names.prepend("Jody")
print(names.pop_start())
I can get the result of Jody. But if instead, I test for
print(names.pop_start() == "Jody")
It shows False. What is the reason?
names.pop_start() returns a Node object. Its data is the string 'Jodie', and because of how you've defined its __str__ method, when you print the node, the string is what you'll see. But the node itself is a node, not a string.
If you compare to the data attribute:
print(names.pop_start().data == "Jody")
...you'll get True, as intended. But it would probably make more sense for pop_start to just return the data anyway, rather than the Node object. Here's how you could do that:
def pop_start(self):
if self.head is None:
return None
else:
data = self.head.data
if self.head is self.tail:
self.tail = None
self.head = self.head.next
return data
not getting expected output. Missing node with data = 6
looks like not insertAfter method properly,
cant find the issue.
pls suggest any other issue too as i am just started with data structures. is there anything need to be kept in mind while studying data structures.
class Node:
def __init__(self,data):
self.data = data
self.next = None
class LinkeList:
def __init__(self):
self.head = None
def push(self,new_data):
new_node = Node(new_data)
new_node.next = self.head
self.head = new_node
def insertAfter(self,prev_node,new_data):
new_node = Node(new_data)
if self.head is None:
self.head = new_node
return
new_node.next = prev_node.next
prev_node.next = new_node
def append(self,new_data):
new_node = Node(new_data)
if self.head is None:
self.head = new_node
return
last = self.head
while last.next != None:
last = last.next
last.next = new_node
def printList(self):
temp = self.head
while temp is not None:
print(temp.data, end = " ")
temp = temp.next
if __name__ =='__main__':
llist = LinkedList()
llist.append(7)
llist.append(8)
llist.push(5)
llist.insertAfter(Node(5),6)
llist.printList()
So the issue is in this line,
llist.insertAfter(Node(5),6)
When you do insertAfter then you need to first get a Node from the current linked list, rather what you are doing is, that you create a new node, then pass it. That new node might have the same value as a . node in your linked list, but it really isn't a part of it, what you need to do is as follows
First implement a get node function, that gets a node from the linked list with the value that you want, something like this
def getNode(self, value):
temp = self.head
while temp is not None:
if temp.value == value:
return temp
else:
temp = temp.next
return None
then do this.
llist.insertAfter(llist.getNode(5),6)
Also put a check for if passed node is not None in insertAfter
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.