fast slow pointers

Floyd's tortoise and hare detects a cycle in O(1) space, but the part that separates strong candidates is the second phase: a short distance argument that pinpoints exactly where the loop begins. Here is the answer and the math.

Two lists that share a tail must intersect at a single node, but matching their lengths without counting is the elegant move interviewers want. Here is the two-pointer trick that makes the lengths cancel out, plus the answer.

The two-pass length-then-delete solution works, but the interview wants a single pass: a gap of N between two pointers, plus a dummy head that makes deleting the real head fall out for free. Here is the clean answer.