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Intuition

The problem requires removing duplicate nodes from a sorted linked list. Since the list is already sorted, duplicate values will always appear consecutively. Therefore, we can traverse the list once, compare the current node with the next node, and remove duplicates as we go.

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Approach

1. Iterate Through the Linked List:

   • Start with the head of the linked list.

   • Use a pointer (current) to traverse the list.

2. Check for Duplicates:

   • While the current node and the next node exist, compare their values:

     • If the value of current.next is equal to current.val, skip the next node by setting current.next to current.next.next.

     • Otherwise, move the current pointer to the next node.

3. Continue Until the End:

   • Repeat the process until all duplicates are removed.

4. Return the Modified List:

   • At the end of the traversal, return the modified linked list starting from head.

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Complexity

• Time Complexity:
  $$O(n)$$
  We traverse the entire linked list once, where (n) is the number of nodes in the list.

• Space Complexity:
  $$O(1)$$
  No additional space is used; the list is modified in place.

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Code

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next

class Solution(object):
    def deleteDuplicates(self, head):
        current = head

        while current:
            while current.next and current.next.val == current.val:
                current.next = current.next.next
            current = current.next

        return head

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Example Test Cases

1. Input: head = [1, 1, 2]
   Output: [1, 2]
   Explanation: The duplicate 1 is removed.

2. Input: head = [1, 1, 2, 3, 3]
   Output: [1, 2, 3]
   Explanation: The duplicates 1 and 3 are removed.

3. Input: head = []
   Output: []
   Explanation: An empty list remains empty.

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