FAQ: Doubly Linked Lists: Python - Removing by Value II

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This community-built FAQ covers the “Removing by Value II” exercise from the lesson “Doubly Linked Lists: Python”.

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FAQs on the exercise Removing by Value II

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2

I find this whole thing fascinating but it quickly becomes hard to follow, is this normal? Should I do everything from the beginning over again till I understand it? Does anyone have any other resources that may help me follow?

1 Like
3

I came up with a different solution and I was wondering if someone could tell me if it is correct.

My solution:

def remove_by_value(self, value_to_remove):
    if value_to_remove == None:
          return None
    elif value_to_remove == self.head_node.get_value():
      self.remove_head()
    elif value_to_remove == self.tail_node.get_value():
      self.remove_tail()
    else:
      current_node = self.head_node
      while current_node:
        next_node = current_node.get_next_node()
        prev_node = current_node.get_prev_node()
        if current_node.get_value() == value_to_remove:
          prev_node.set_next_node(next_node)
          next_node.set_prev_node(prev_node)
          return current_node
        current_node = next_node

Given solution:

def remove_by_value(self, value_to_remove):
    node_to_remove = None
    current_node = self.head_node

    while current_node != None:
      if current_node.get_value() == value_to_remove:
        node_to_remove = current_node
        break

      current_node = current_node.get_next_node()

    if node_to_remove == None:
      return None

    if node_to_remove == self.head_node:
      self.remove_head()
    elif node_to_remove == self.tail_node:
      self.remove_tail()
    else:
      next_node = node_to_remove.get_next_node()
      prev_node = node_to_remove.get_prev_node()
      next_node.set_prev_node(prev_node)
      prev_node.set_next_node(next_node)

    return node_to_remove
2 Likes
4

I had a question about removing by value II in Double Linked Lists, for the node to be removed we dereference it by:
[python]
next_node = node_to_remove.get_next_node()
prev_node = node_to_remove.get_prev_node()
next_node.set_prev_node(prev_node)
prev_node.set_next_node(next_node)
[/python]
Here node_to_remove is the node containing the value to be removed. We change the links of it’s next node and previous node to link each other, but there is no mention of dereferencing node_to_remove’s next_node and prev_node. Like:
[python]
node_to_remove.set_next_node(None);
node_to_remove.set_prev_node(None);
[/python]
I can understand why it’s unnecessary coz when we traverse the double-linked list, it will skip over node_to_remove. But, I was wondering, would it still exist in the computer memory?? If we had to remove multiple values in b/w wouldn’t it be a problem??

5

Works and I like your implementation way better!

6 
def remove_by_value(self, value_to_remove):
    node_to_remove = None
    current_node = self.head_node
    while current_node != None:
      if current_node.get_value() == value_to_remove:
        node_to_remove = current_node
        break
    current_node = current_node.get_next_node()
    if node_to_remove == None:
      return None
    if node_to_remove == self.head_node:
      self.remove_head()
#added code for step 2 below
    elif node_to_remove == self.tail_node:
      self.remove_tail()

i hit run on step 2 and the button turns into a spinning wheel. waiting, waiting, waiting, then:
Failed to test your code.

7

i delete the code and it still fails, delete more same result, what is wrong?

8

wow, i didn’t see the fact that the while was infinite, human error