delete_first

delete_first.py

def delete_first(self):
    deleted = self.head
    if self.head:
        self.head = self.head.next
        deleted.next = None
    return deleted

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😊

Lol @parvinderandroid

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

Great solution

…

On Fri, 12 Jan 2024, 00:27 lator, ***@***.***> wrote: ***@***.**** commented on this gist. ------------------------------ 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 — Reply to this email directly, view it on GitHub <https://gist.github.com/adarsh0806/02d8e1d54d510294e75dfbc0d9bd7059#gistcomment-4827398> or unsubscribe <https://github.com/notifications/unsubscribe-auth/AWKPMQYFPAPCNK72KSSDLNDYOBYPXBFKMF2HI4TJMJ2XIZLTSKBKK5TBNR2WLJDUOJ2WLJDOMFWWLO3UNBZGKYLEL5YGC4TUNFRWS4DBNZ2F6YLDORUXM2LUPGBKK5TBNR2WLJDHNFZXJJDOMFWWLK3UNBZGKYLEL52HS4DFVRZXKYTKMVRXIX3UPFYGLK2HNFZXIQ3PNVWWK3TUUZ2G64DJMNZZDAVEOR4XAZNEM5UXG5FFOZQWY5LFVAZTMMZRHEYDENVHORZGSZ3HMVZKMY3SMVQXIZI> . You are receiving this email because you commented on the thread. Triage notifications on the go with GitHub Mobile for iOS <https://apps.apple.com/app/apple-store/id1477376905?ct=notification-email&mt=8&pt=524675> or Android <https://play.google.com/store/apps/details?id=com.github.android&referrer=utm_campaign%3Dnotification-email%26utm_medium%3Demail%26utm_source%3Dgithub> .