Awesome tutorial as always! I added one more condition at the end as I was not able to remove the key if the circular linked list contained only one node. Hope this is right. if self.head==self.head.next and self.head.data==key: self.head=None
def remove(self,key): if self.head.data == key: cur = self.head while cur.next != self.head: cur = cur.next cur.next = self.head.next self.head = self.head.next else: cur = self.head prev = None while cur.data != key: prev = cur cur = cur.next prev.next = cur.next I used a similar method but did a little bit change in the while loop in the else statement. I think it works! As always, thank you for the tutorial and keep on the great work!!
Hi Vincent ! thanks for the great tutorial! 3 issues 1, cur=cur.next is inefficient, if key is at tail, the while loop will end up running *twice* before it can exit out. I suggest use break 2, throw error when it's an empty list which can be easily modified by adding a if statement 3, cant remove when there is only one node, someone else in the comment section already pointed it out and gave great solution thanks again for your work!
I think the clarity of less code for the one edge case of just an extra iteration is probably worth it. Thanks for pointing that out, and thanks for watching!
Salute to you. You are really awesome. Just one thing i wanted to point in above code. If CL is A>B>C>C>D, after removing C, it will result A>B>C>D, but if CL is A>B>C>D>C, after removing C, it will give A>B>D
i am really learning a lot from all of your data structure playlist videos. i implemented my own version of the remove method. seems to work fine for all the edge cases. def remove(self, data): head = node = prev = self.head while node.next != head: if node.data == data and node != head: break prev, node = node, node.next if self.head.data == data: self.head = head.next node.next = self.head if node.data == data: prev.next = node.next
Thank you, Sarthak! My recommendation to you would be to try to take it out and see what happens for some edge cases. Go ahead and also pepper in some print statements to see what occurs and see if you can figure it out from that. Let me know if not. Cheers!
Great video and the visual representation in the previous one really helped. I still do have one question; how do we delete the last node in a list? I get that this would be the case if the next-pointer of a node would point to itself, but how do we let the __repr__ or print function know? Is it enough if we set the data of that last node to None?
Thank you. In order to delete the last node, you would check if the next node is head. If the next node is head, you know you're at the last node. Does that make sense?
Hello, @LucidProgramming I have a question regarding the deletion of the node like you wrote while curr.next != self.head: prev = curr curr = curr.next #why did you move the next pointer here if curr.data is data: prev.next = curr.next curr = curr.next #why not here, like we used to do in earlier videos please explain I am bit confused. Thank you [UPDATE] * We can write the above program like this also. This also works flawlessly.* else: curr = self.head prev = None while curr.next != self.head: nxt = curr.next if curr.data is data: prev.next = nxt prev = curr curr = nxt Thank you
@@LucidProgramming With my update i am telling that we can write the program in that way also. Yes, I got the answer of my question by myself. It was the way of writing the program, I wrote slightly different from yours. Thank you
the github link in the video description points to the split list repo. here is the remove repo if anyone is looking for it. github.com/vprusso/youtube_tutorials/blob/master/data_structures/linked_list/circular_linked_list/circular_linked_list_remove.py
For else condition, the more efficient way is to check the key in while loop. This would ensure less iteration unless you want to delete the last node. Let me know if you think otherwise.
Hey so appreciate your detailed tutorial ! I wanna ask if it would be an easy way to write: else: cur = self.head prev = None While cur.data != key: prev = cur cur = cur.next prev.next = cur.next
Hello! Thanks for watching and for the comment. If you run through your suggestion, I think you'll encounter an error if the element you're looking for is not present in the list. Basically, you keep looking while the data is not equal to the key. Well, since this is a circular linked list and say the element is not in the list, you'll just keep looping. Did you try to make these changes yourself? I assume this is what you would have seen if you did so.
LucidProgramming so grateful for your patient reply and instruction! You're right, I just ignore the scenario that if the element is not in the list. I only consider the case that all the element I wanna remove! Thanks again for your reply :)
Hey, I am the one who asked why self.head = self.head.next is a must. I have made comment under another video, sorry for that! Could you please explain line 91 for me ?
Could we just break out of the loop once we have found the key and altered the next position of the previous nose as below... As it is not required to go through the list again since we aimed at removing a single node.. code: def remove(self,key): if self.head.data == key: cur = self.head while cur.next != self.head: cur = cur.next cur.next = self.head.next self.head = self.head.next else: prev = None cur = self.head while cur.next != self.head: prev = cur cur = cur.next if cur.data == key: prev.next = cur.next #cur = cur.next break
![โรงเรียนตอนตี 2 โคตรหลอน..!! [โรงเรียนไดเม่]](http://i.ytimg.com/vi/2MAEMwqVYh8/mqdefault.jpg)
Awesome tutorial as always!
I added one more condition at the end as I was not able to remove the key if the circular linked list contained only one node.
Hope this is right.
if self.head==self.head.next and self.head.data==key:
self.head=None
Hmm, I thought this case was properly dealt with, but perhaps I missed it. Thanks for the catch, and appreciate you watching and commenting! Cheers :)
After doing this you also have to add:
if self.head is None:
return
at beginning of the method.
def remove(self,key):
if self.head.data == key:
cur = self.head
while cur.next != self.head:
cur = cur.next
cur.next = self.head.next
self.head = self.head.next
else:
cur = self.head
prev = None
while cur.data != key:
prev = cur
cur = cur.next
prev.next = cur.next
I used a similar method but did a little bit change in the while loop in the else statement. I think it works!
As always, thank you for the tutorial and keep on the great work!!
Hi Vincent ! thanks for the great tutorial!
3 issues
1, cur=cur.next is inefficient, if key is at tail, the while loop will end up running *twice* before it can exit out. I suggest use break
2, throw error when it's an empty list which can be easily modified by adding a if statement
3, cant remove when there is only one node, someone else in the comment section already pointed it out and gave great solution
thanks again for your work!
I think the clarity of less code for the one edge case of just an extra iteration is probably worth it. Thanks for pointing that out, and thanks for watching!
Salute to you. You are really awesome.
Just one thing i wanted to point in above code.
If CL is A>B>C>C>D, after removing C, it will result A>B>C>D, but if CL is A>B>C>D>C, after removing C, it will give A>B>D
Thanks for the comment, and thanks for watching!
This playlist is amazing, thank you for your amazing work!!
Thank you--much appreciated!
i am really learning a lot from all of your data structure playlist videos. i implemented my own version of the remove method. seems to work fine for all the edge cases.
def remove(self, data):
head = node = prev = self.head
while node.next != head:
if node.data == data and node != head: break
prev, node = node, node.next
if self.head.data == data:
self.head = head.next
node.next = self.head
if node.data == data: prev.next = node.next
Awesome good to hear, and thanks for sharing!
Great video.Why is there a need for " cur = cur.next" in the last line of the function?The program seems to run without it
Thank you, Sarthak! My recommendation to you would be to try to take it out and see what happens for some edge cases. Go ahead and also pepper in some print statements to see what occurs and see if you can figure it out from that. Let me know if not. Cheers!
@Daniel Kua What are "all the edge cases" in this instance?
@Daniel Kua Hahaha, so you're saying you tried nothing when you tried all the edge cases?
@Daniel Kua Google. It's a wonderful thing.
Great video and the visual representation in the previous one really helped. I still do have one question; how do we delete the last node in a list? I get that this would be the case if the next-pointer of a node would point to itself, but how do we let the __repr__ or print function know? Is it enough if we set the data of that last node to None?
Thank you. In order to delete the last node, you would check if the next node is head. If the next node is head, you know you're at the last node. Does that make sense?
@@LucidProgramming yes it does! Checking if the next node is head and then setting not it's data field but the node itself to None worked for me
@@tomio444 Great. Cheers for watching and thanks for letting me know!
Hello, @LucidProgramming I have a question regarding the deletion of the node like you wrote
while curr.next != self.head:
prev = curr
curr = curr.next #why did you move the next pointer here
if curr.data is data:
prev.next = curr.next
curr = curr.next
#why not here, like we used to do in earlier videos
please explain I am bit confused.
Thank you
[UPDATE]
* We can write the above program like this also. This also works flawlessly.*
else:
curr = self.head
prev = None
while curr.next != self.head:
nxt = curr.next
if curr.data is data:
prev.next = nxt
prev = curr
curr = nxt
Thank you
With the update, did you answer your own question? I can't quite tell if you did or not. Cheers, and thanks again for your comments!
@@LucidProgramming With my update i am telling that we can write the program in that way also.
Yes, I got the answer of my question by myself. It was the way of writing the program, I wrote slightly different from yours.
Thank you
the github link in the video description points to the split list repo. here is the remove repo if anyone is looking for it.
github.com/vprusso/youtube_tutorials/blob/master/data_structures/linked_list/circular_linked_list/circular_linked_list_remove.py
For else condition, the more efficient way is to check the key in while loop. This would ensure less iteration unless you want to delete the last node. Let me know if you think otherwise.
Hi Dev. I'm not sure I agree with your assessment. Your approach would involve adding an unnecessary check on every iteration of the loop.
Hey so appreciate your detailed tutorial !
I wanna ask if it would be an easy way to write:
else:
cur = self.head
prev = None
While cur.data != key:
prev = cur
cur = cur.next
prev.next = cur.next
Hello! Thanks for watching and for the comment. If you run through your suggestion, I think you'll encounter an error if the element you're looking for is not present in the list. Basically, you keep looking while the data is not equal to the key. Well, since this is a circular linked list and say the element is not in the list, you'll just keep looping. Did you try to make these changes yourself? I assume this is what you would have seen if you did so.
LucidProgramming
so grateful for your patient reply and instruction! You're right, I just ignore the scenario that if the element is not in the list. I only consider the case that all the element I wanna remove! Thanks again for your reply :)
No worries, that's an easy case to miss. Thanks for watching, and happy coding!
Hey, I am the one who asked why self.head = self.head.next is a must. I have made comment under another video, sorry for that!
Could you please explain line 91 for me ?
I would suggest you try taking the line out and trying some edge cases to see why this is there.
Could we just break out of the loop once we have found the key and altered the next position of the previous nose as below...
As it is not required to go through the list again since we aimed at removing a single node..
code:
def remove(self,key):
if self.head.data == key:
cur = self.head
while cur.next != self.head:
cur = cur.next
cur.next = self.head.next
self.head = self.head.next
else:
prev = None
cur = self.head
while cur.next != self.head:
prev = cur
cur = cur.next
if cur.data == key:
prev.next = cur.next
#cur = cur.next
break
Sure, that would save you some potential checks.
does this work if head is the only node and its pointing to itself?
Have you tried that case?
LucidProgramming I wouldn’t have asked if I did, but thanks for the answer that was really helpful
@@jamesjanes7936 Cheers, and thanks for watching!
Last node remove isn't working
Seems to work for me. Are you sure you're using the code on the GitHub?