Traverse the list once with two pointers to locate the first two nodes with value v, then reverse the sublist between them using standard pointer manipulation. Handle edge cases like adjacent nodes or fewer than two matches by returning the list unchanged.
Pro tip: Clarify whether the target nodes themselves should be reversed or only the nodes strictly between them; explicitly state your assumption and handle the edge case where A and B are adjacent (empty sublist).
Confirm that only nodes strictly between A and B are reversed, and that if fewer than two nodes with value v exist, the list is unchanged. Consider cases like empty list, single node, adjacent matches, and all nodes matching.
Traverse the list with a single pointer, counting occurrences of value v. Record the first node (A) and the second node (B) when found. If a second match is not found, return the original list.
Use three pointers (prev, curr, next) to reverse the nodes strictly between A and B. Ensure A.next points to the new head of the reversed sublist and the tail of the reversed sublist points to B.
After reversal, reconnect the reversed sublist properly: A.next should point to the new first node of the reversed portion, and the last node of the reversed portion should point to B. Return the head of the modified list.
State that the solution runs in O(n) time and O(1) extra space. Walk through a few test cases (e.g., matches at head/tail, adjacent matches) to verify correctness.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.