Remove Duplicates from Sorted List II

To solve the Remove Duplicates from Sorted List II problem, I would first clarify that we need to delete every node that has a duplicate value, leaving only numbers that appear exactly once. Since the input is sorted, identical values are adjacent, which simplifies detection. My approach involves using a dummy node to handle edge cases where the head itself might be a duplicate. I'll initialize a pointer called 'prev' to this dummy node and another pointer 'curr' starting at the head. As I iterate through the list, I check if curr.next exists and if their values match. If they do, I know curr and potentially more nodes are duplicates. I will advance curr while its value equals the next node's value, effectively skipping all instances of that number. Then, I set prev.next to curr.next, bypassing the entire block of duplicates. If the values don't match, it means curr is unique, so I simply move prev to curr and advance curr. This ensures O(n) time complexity with O(1) space. For example, given 1->2->3->3->4->4->5, the algorithm skips both 3s and both 4s, returning 1->2->5. This method avoids special casing the head node and aligns with Amazon's preference for robust, bug-free solutions that handle boundary conditions gracefully.