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Remove Duplicates from Sorted List LC #83 Easy Linked List · In-Place
Problem

Given the head of a sorted linked list, delete all duplicates so that each element appears only once. Return the modified list head (still sorted).

Example 1
Input: head = [1,1,2]
Output: [1,2]
Example 2
Input: head = [1,1,2,3,3]
Output: [1,2,3]
Constraints: 0 ≤ list length ≤ 300  |  −100 ≤ Node.val ≤ 100  |  List is sorted ascending
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Custom:
List Visualization
Input list (modified in-place)
Result list (unique nodes)
Variables
curr.val
—
curr.next.val
—
Action
—
Step Logic
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Algorithm
1
Set curr = head.
2
Loop while curr != null && curr.next != null.
3
Duplicate: curr.val == curr.next.val → curr.next = curr.next.next. Stay at curr.
4
Unique: advance curr = curr.next.
5
Return head.
Time
O(n)
Space
O(1)
Key Insight

Because the list is already sorted, all duplicates of a value are adjacent. We rewire curr.next to skip over the duplicate node — O(1) pointer update, no extra memory. The skip arc shows exactly which pointer is redirected. curr never moves backward; it either stays (if the next node was a duplicate that got unlinked) or advances forward.