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| def delete_first(self): | |
| deleted = self.head | |
| if self.head: | |
| self.head = self.head.next | |
| deleted.next = None | |
| return deleted |
@ScottDWebster
Here is the reason you "Return" the value of the Deleted Element: Its to implement POP method, if you are implementing Stacks.
Along with Deleting the Head, POP is also supposed to return its value... hence the "return" statement. Its "future-proofing" your "delete-first" to support 'POP' functionality.
def delete_first(self):
if self.head:
deleted_item = self.head
self.head = deleted_item.next
deleted_item.next = None
return deleted_element
else:
return None
def delete_first(self):
"Delete the first (head) element in the LinkedList as return it"
deleted = self.head
self.head = self.head.next if self.head else None
return deleted
It is the same with the first answer. But the if statement much smaller.
In that code:
def delete_first(self):
deleted = self.head
if self.head:
self.head = self.head.next
deleted.next = None
return deleted
is the deleted.next = None line necessary? Is it required for python garbage collector?
Also, I have a doubt that whenever we are pushing or poping, we are not actually deleting something, is it efficient or python garbage collector smart enough to reuse them
@epsi95 by making deleted.next = None you prevent the possibility to someone iterate / modify the remaining elements of the stack.
@plicatibu ya, you are right, actually, I didn't think from user point of view.
This works well. But from Udacity perspective. It doesn't specify you need to return anything on the pop command. It can just work without any return right?
Another way of writing it.
if self.head:
deleted_element = self.head
self.head = deleted_element.next
deleted_element.next = None
return deleted_element
else:
return self.head
Another way of writing it.
if self.head: deleted_element = self.head self.head = deleted_element.next deleted_element.next = None return deleted_element else: return self.head `why there is no function please tell
he's pasting just the body of the function it's not a solution without function definition. :P
Another sweet and succinct way is:-
def delete_first(self):
"Delete the first (head) element in the LinkedList and return it"
current = self.head
if current:
self.head = current.next
return current
This is another way to achieve the same result!
def delete_first(self): elementToRemove = None if(self.head != None ): elementToRemove = self.head self.head = self.head.next return elementToRemove
Did not expect to see you here😊
Does this not work?
def delete_first(self):
"Delete the first (head) element in the LinkedList as return it"
if self.head:
head_ele = self.head
self.head = head_ele.next
head_ele.next = None
return head_ele.value
else:
return None
def delete_first(self):
elementToRemove = None
if(self.head != None ):
elementToRemove = self.head
self.head = self.head.next
return elementToRemove
Not memory efficient, since another Element is instantiated, yet it takes less lines
def delete_first(self):
head_element = Element(self.head.value)
if self.head is not None:
self.head = self.head.next
return head_element
def delete_first(self):
current = self.head
if self.head:
self.head=self.head.next
return current
def delete_first(self):
if not self.head:
return None
else:
x = self.head
self.head = self.head.next
return x
Here's what I wrote
def delete_first(self):
"Delete the first (head) element in the LinkedList as return it"
current = self.head
if current and current.next != None:
self.head = current.next
else:
self.head = None
return current
This is how I implemented it :)
def delete_first(self):
"Delete the first (head) element in the LinkedList as return it"
if not self.head:
return None
if self.head:
deleted = self.head
self.head = self.head.next
return deleted.value
Here's another version
def delete_first(self):
"Delete the first (head) element in the LinkedList as return it"
self.head, node = (self.head.next, Element(self.head.value)) if self.head else (None, None)
return node@yaakash @karthik19894 I think it will be automatically garbage collected, If we see logically since when self.head is returned assigning to deleted variable the reference counter will be decreased after returning from delete_first if I am not wrong the reference counter would be 0 if the return value is not reassigned then garbage collector knows that if can clean the data since nothing is pointing to it.
def delete_first(self):
deleted = self.head
if self.head:
self.head = self.head.next
deleted.next = None
return deleted
My implementation:
def delete_first(self):
"Delete the first (head) element in the LinkedList as return it"
if self.head and self.head.next:
deleted_element = self.head
self.head = self.head.next
return deleted_element
elif self.head and self.head.next == None:
deleted_element = self.head
self.head = None
return deleted_element
else:
return None
My implementation:
def delete_first(self):
"Delete the first (head) element in the LinkedList as return it"
node = self.head
try: self.head = self.head.next
except AttributeError: pass
finally: return node
This is according to the official best practices of Python, adhering to the EAFP. Basically claiming that it's computationally faster (in no-error cases) to use try/except instead of if/else statements.
My Implementation:
def delete_first(self):
"Delete the first (head) element in the LinkedList as return it"
if self.head is None:
return None
current = self.head
self.head = self.head.next
return current
I hope this helps
"Delete the first (head) element in the LinkedList as return it"
temp = self.head
self.head = temp.next if temp else None
return temp```
Hope this helps
def delete_first(self):
deleted_element = None
if self.head is not None:
deleted_element = self.head
deleted_element.next = None
if self.head.next is not None:
self.head = self.head.next
else:
self.head = None
return deleted_element
Compared to most solution, I think I was too explicit
def delete_first(self):
if self.head:
old_head = self.head
if self.head.next:
new_head = self.head.next
self.head = new_head
old_head.next = None
else:
self.head = None
return old_head
return None
Echoing @ScottDWebster... why "return" anything?
Is that Garbage Collection in Python?