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dsa-solutions/lc-solutions/0000-0099/0082-remove-duplicates-from-sorted-list-2.md

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+---
+id: remove-duplicates-from-sorted-list-2
+title: Remove Duplicates from Sorted List II (Leetcode)
+sidebar_label: 0082-RemoveDuplicatesFromSortedListII
+description: Given the head of a sorted linked list, delete all nodes that have duplicate numbers, leaving only distinct numbers from the original list. Return the linked list sorted as well.
+---
+
+## Problem Description
+
+| Problem Statement | Solution Link | LeetCode Profile |
+| :---------------- | :------------ | :--------------- |
+| [Remove Duplicates from Sorted List II](https://leetcode.com/problems/remove-duplicates-from-sorted-list-ii/) | [Merge Two Sorted Lists Solution on LeetCode](https://leetcode.com/problems/remove-duplicates-from-sorted-list-ii/solutions) |  [Aaradhya Singh ](https://leetcode.com/u/keira_09/) |
+
+
+## Problem Description
+
+Given the head of a sorted linked list, delete all nodes that have duplicate numbers, leaving only distinct numbers from the original list. Return the linked list sorted as well.
+
+### Examples
+
+#### Example 1
+
+- **Input:** $head = [1,2,3,3,4,4,5]$
+- **Output:** $[1,2,5]$
+
+
+#### Example 2
+
+- **Input:** $head = [1,1,1,2,3]$
+- **Output:** $[2,3]$
+
+
+
+### Constraints
+
+- The number of nodes in the list is in the range [0, 300].
+- $-100 <= Node.val <= 100$
+- The list is guaranteed to be sorted in ascending order.
+
+
+### Intuition
+
+
+The goal is to remove all elements from a sorted linked list that have duplicate values, ensuring that each element appears only once. We use a dummy node to simplify handling edge cases and traverse the list, removing duplicates as we encounter them.
+
+
+### Approach
+
+1. **Initialization:**
+
+    - Create a dummy node to handle edge cases where the head itself might be a duplicate.
+    - Initialize two pointers, prev (starting at dummy) and curr (starting at head).
+
+2. **Traversal and Duplicate Removal:**
+
+    - Traverse the linked list using the curr pointer.
+    - For each node, check if the current value matches the next node's value.
+    - If duplicates are detected, use a loop to skip all nodes with that value, deleting them.
+    - Update the prev pointer's next to point to the first non-duplicate node after the series of duplicates.
+    - If no duplicates are found, move both prev and curr pointers forward.
+
+3. **Completion:**
+
+    - After the loop, update the head to point to the node after the dummy.
+    - Delete the dummy node to free memory.
+    Return the updated head of the linked list.
+
+### Solution Code
+
+#### Python
+
+```py
+class ListNode:
+    def __init__(self, val=0, next=None):
+        self.val = val
+        self.next = next
+
+class Solution:
+    def deleteDuplicates(self, head: ListNode) -> ListNode:
+        if not head or not head.next:
+            return head
+
+        dummy = ListNode(0)  # Dummy node to handle the case when head is a duplicate
+        dummy.next = head
+
+        prev = dummy
+        curr = head
+
+        while curr and curr.next:
+            if prev.next.val == curr.next.val:
+                val = curr.next.val
+                while curr and curr.val == val:
+                    temp = curr
+                    curr = curr.next
+                    del temp  # Marking the node for garbage collection
+                prev.next = curr
+            else:
+                prev = prev.next
+                curr = curr.next
+
+        head = dummy.next  # Update head in case it has changed
+        del dummy  # Marking the dummy node for garbage collection
+        return head
+```
+
+#### Java
+
+```java
+class ListNode {
+    int val;
+    ListNode next;
+    ListNode() {}
+    ListNode(int val) { this.val = val; }
+    ListNode(int val, ListNode next) { this.val = val; this.next = next; }
+}
+
+class Solution {
+    public ListNode deleteDuplicates(ListNode head) {
+        if (head == null || head.next == null) {
+            return head;
+        }
+
+        ListNode dummy = new ListNode(0);  // Dummy node to handle the case when head is a duplicate
+        dummy.next = head;
+
+        ListNode prev = dummy;
+        ListNode curr = head;
+
+        while (curr != null && curr.next != null) {
+            if (prev.next.val == curr.next.val) {
+                int val = curr.next.val;
+                while (curr != null && curr.val == val) {
+                    ListNode temp = curr;
+                    curr = curr.next;
+                    temp = null;  // Marking the node for garbage collection
+                }
+                prev.next = curr;
+            } else {
+                prev = prev.next;
+                curr = curr.next;
+            }
+        }
+
+        head = dummy.next;  // Update head in case it has changed
+        dummy = null;  // Marking the dummy node for garbage collection
+        return head;
+    }
+}
+```
+
+#### C++
+
+```cpp
+class Solution {
+public:
+    ListNode* deleteDuplicates(ListNode* head) {
+        if (head == nullptr || head->next == nullptr) {
+            return head;
+        }
+
+        ListNode* dummy = new ListNode(0);  // Dummy node to handle the case when head is a duplicate
+        dummy->next = head;
+
+        ListNode* prev = dummy;
+        ListNode* curr = head;
+
+        while (curr != nullptr && curr->next != nullptr) 
+        {
+            if (prev->next->val == curr->next->val) 
+            {
+                int val = curr->next->val;
+                while (curr != nullptr && curr->val == val) 
+                {
+                    ListNode* temp = curr;
+                    curr = curr->next;
+                    delete temp;
+                }
+                prev->next = curr;
+            } else {
+                prev = prev->next;
+                curr = curr->next;
+            }
+        }
+
+        head = dummy->next;  // Update head in case it has changed
+        delete dummy;  // Delete the dummy node
+        return head;
+    }
+};
+```
+
+### Conclusion
+
+The provided code effectively removes duplicates from a sorted linked list by iterating through the list and adjusting the pointers accordingly to skip duplicate nodes. It uses a dummy node to handle cases where the head itself is a duplicate and performs the deletion in place without modifying the values within the nodes. The solution has a time complexity of $O(n)$, where n is the number of nodes in the linked list, due to the linear traversal required to identify and remove duplicates. The space complexity is $O(1)$ since the algorithm operates in constant space, only using a few pointers and temporary variables regardless of the input size. Overall, this solution offers an efficient and straightforward approach to handling duplicate removal in a sorted linked list.