Remove Duplicates in Sorted Linked List

In this task, I worked on removing duplicate elements from a sorted linked list.

Since the list is already sorted, duplicate values will always appear next to each other.

What I Did

I created a function removeDuplicates that:

• Takes the head of a linked list
• Removes duplicate nodes
• Returns the modified list with only unique elements

How I Solved It

Instead of using extra space , I used a simple pointer traversal.

Approach

I used one pointer:

• curr → starts from the head and traverses the list

Logic Behind It

At each step:

• Compare the current node with the next node

• If both values are the same:

  • Skip the next node → curr.next = curr.next.next

• Otherwise:

  • Move to the next node → curr = curr.next

Code

class ListNode:
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next

def removeDuplicates(head):
    curr = head

    while curr and curr.next:
        if curr.val == curr.next.val:
            curr.next = curr.next.next  
        else:
            curr = curr.next

    return head

How It Works

Because the list is sorted:

• All duplicates are adjacent
• So we only need to compare neighboring nodes

When a duplicate is found, we simply bypass it, effectively removing it from the list.

Time Complexity

• O(n) → traverse the list once

Space Complexity

• O(1) → no extra space used

Example

Original:
1 → 1 → 2 → 3 → 3 → 4 → None

After removing duplicates:
1 → 2 → 3 → 4 → None