# 🧩 Remove Duplicates from Sorted Linked List (Java)

🚀 Problem Statement

You are given the head of a sorted singly linked list.
Your task is to:

• Remove all duplicate nodes
• Keep only one occurrence of each value
• Do it in-place (no extra space)

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💡 Key Idea

Since the list is already sorted, duplicates will always be adjacent.

👉 So we just:

• Traverse the list
• Compare current node with next node
• If equal → skip the next node

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🧠 Approach

1. Start from the head
2. While current node and next node exist:

• If current.data == current.next.data

  • Remove next node → current.next = current.next.next

• Else move forward

  1. Return the head

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🛠️ Java Implementation (GFG Style)

```java id="r8p4kx"
class Solution {
// Function to remove duplicates from sorted linked list
Node removeDuplicates(Node head) {
// Base case
if (head == null) return null;

    Node current = head;

    while (current != null && current.next != null) {
        if (current.data == current.next.data) {
            // Skip duplicate node
            current.next = current.next.next;
        } else {
            current = current.next;
        }
    }

    return head;
}

}

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## 🔍 Example



```text id="o4w6f3"
Input:
1 -> 1 -> 2 -> 3 -> 3

Output:
1 -> 2 -> 3

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⏱️ Complexity Analysis

Type	Complexity
Time	O(n)
Space	O(1)

• We traverse the list once
• No extra memory used

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⚠️ Common Mistakes

❌ Moving pointer after deleting (can skip nodes)
❌ Using extra data structures (not needed here)
❌ Forgetting null checks

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📌 Key Takeaways

• Sorted property makes the problem easy
• Just compare adjacent nodes
• Modify links instead of creating new nodes

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🎯 Final Thoughts

This is a simple but important linked list problem. It teaches:

• Pointer manipulation
• In-place operations
• Efficient traversal

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