FAQ: Technical Interview Problems in Python: Linked Lists - Remove Duplicates

Frequently Asked Questions Python FAQ
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This community-built FAQ covers the “Remove Duplicates” exercise from the lesson “Technical Interview Problems in Python: Linked Lists”.

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This exercise can be found in the following Codecademy content:

Technical Interview Practice: Python

FAQs on the exercise Remove Duplicates

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2

Not sure what’s going on but this page (lesson) seems locked up. When I click on Save/Run the icon just keeps spinning and never finishes. This has been going on now for 24 hours.

4

It does appear to be the script of mine linked_list.py. I would first suspect the while loop. I’m just not sure why its hanging.

	def remove_duplicates(self):
		#Store a reference to the head node
		current_node = self.head

		#Check if there a .next node an that .next node contain the same values as the current node
		# remove the duplicate by reassigning the .next property of the current node.
		while current_node:
			while current_node.next and current_node.next.val == current_node.val:
				current_node.next = current_node.next.next

		current_node = current_node.next
		return self

EDIT
Indention got me, fixed.

	def remove_duplicates(self):
		#Store a reference to the head node
		current_node = self.head

		#Check if there a .next node an that .next node contain the same values as the current node
		# remove the duplicate by reassigning the .next property of the current node.
		while current_node:
			while current_node.next and current_node.next.val == current_node.val:
				current_node.next = current_node.next.next
			current_node = current_node.next
		return self
5

Hi,

hat happen because of an infinite loop probably… have to refresh the browser and fix the code.

I created a different solution to remove duplicated using only one loop:

My solution 
def remove_duplicates(self):
    previous_node = self.head
    current_node = self.head.next
    
    # traverse to the end:
    while current_node:
      # check if the value is unique:
      value = current_node.val
      if value == previous_node.val:
        # this value is duplicated; remove node:
        previous_node.next = current_node.next
      else:
        # unique value; refresh the previous node
        previous_node = current_node
      
      # move to the next node:
      current_node = current_node.next
      
    return self
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