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@MadPhysicist "It [deque] behaves like a linked list in almost every way, even if the name is different." — it is either wrong or meaningless: it is wrong because linked lists may provide different guarantees for time complexities e.g., you can remove an element (known position) from a linked list in O(1) while deque doesn't promise it (it is O(n)).
@Kr0e I wouldn't say iterating over the list is as slow as linked lists, but that iterating over the values of the linked lists is a slow as a linked list, with the caveat that Fred mentioned. For example, iterating over a list to copy it to another one should be faster than a linked list.
itr2 = ll2.head. for i in range(1,n2): temp = Node(l2[i]) itr2.next = temp. itr2 = itr2.next. Merging two sorted linked list using merge function by passing the head of the two linked list. itr = merge(ll1.head,ll2.head) "merge" function returns an iterator itself whose values are printed as: while itr != None:
5. I would like to create a delete_node function that deletes the node at the location in the list as a count from the first node. So far this is the code I have: class node: def __init__(self): self.data = None # contains the data. self.next = None # contains the reference to the next node. class linked_list:
To recursively print the list, you'll want to have two functions, one to handle the wrapper and a helper function to do the recursive processing of the nodes. def print_list(linked_list): # Non-recursive outer function. You might want. _print_list_helper(linked_list.first) # to update it to handle empty lists nicely!
I'm a beginner in python and just learning linked lists so bear with me if this seems like a stupid question. I have a linked list of nodes where each node is an instance of a class. How do I retrieve the node in the linked list with a certain instance attribute = x? I have a linked list of nodes where each node is a dictionary.
I created a linked list in python, and I created a pointer variable pointing to the same node. However, when I try to update the pointer, the linked list doesn't update. It only updates when I use the orignal notation. Here are the three lines of concern, followed by the code snippet:
class Node: ''' Class to create Node and next part of the linked list ''' def __init__(self,data): self.data = data self.next = None def createLL(arr): ''' linked list creation ''' head = Node(arr[0]) tail = head for data in arr[1:]: tail.next = Node(data) tail = tail.next return head def midPointOfLL(head): ''' Here, i am using two pointers ...
Reading the abstract/prototype, it looks like the creator of the problem wanted to solve this with recursive/dynamic programming methodology. This is a pretty standard recursive algorithm introduction. It's more about understanding how to write elegant recursive code more than creating linked-list in Python (not really useful or common).
The way to do this is to change what your pointer (val) is pointing at, and the thing that you want it to point at next is the next node along. Therefore: val = ll.head. while val is not None: # do something with val here. val = val.next. edited Dec 1, 2016 at 17:28. answered Dec 1, 2016 at 17:21. wildwilhelm.