Given the head of a singly linked list and two 1-indexed positions left and right (with left ≤ right), reverse only the nodes from position left to position right, then return the head of the modified list. Nodes outside that window keep their original order. The single-pass solution leans on a trailing prev pointerTrailing Prev PointerKeep a pointer one step behind the section being modified. Because reversing a run of nodes only ever changes .next pointers — never values — anchoring prev right before the window means every node pulled out of the window can be re-inserted immediately after prev with a fixed, three-line relink, with no lookahead and no extra memory.: once prev is parked right before the window, the window's own first node (curr) never moves — it's exactly where it needs to end up as the window's last node — and each subsequent node just gets unhooked and re-inserted right after prev, one at a time, until the whole window is reversed.