Question

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:
    def __init__(self):
        self.cur_node = None

    def add_node(self, data):
        new_node = node() # create a new node
        new_node.data = data
        new_node.next = self.cur_node # link the new node to the 'previous' node.
        self.cur_node = new_node #  set the current node to the new one.

    def list_print(self):
        node = ll.cur_node
        while node:
            print node.data
            node = node.next
    def delete_node(self,location):
        node = ll.cur_node
        count = 0
        while count != location:
            node = node.next
            count+=1
        delete node


ll = linked_list()
ll.add_node(1)
ll.add_node(2)
ll.add_node(3)

ll.list_print()

Answer 1

You shouldn’t literally delete a node in Python. If nothing points to the node (or more precisely in Python, nothing references it) it will be eventually destroyed by the virtual machine anyway.

If n is a node and it has a .next field, then:

n.next = n.next.next 

Effectively discards n.next, making the .next field of n point to n.next.next instead. If n is the node before the one you want to delete, this amounts to deleting it in Python.

[P.S. the last paragraph may be a bit confusing until you sketch it on paper – it should then become very clear]