Floyd's tortoise-and-hare detects a cycle in O(n) time and O(1) space using two pointers moving at different speeds.
Detection
Advance slow by one node and fast by two on each step. If the list ends (fast or fast.next becomes null), there is no cycle. If there is a cycle, the fast pointer laps the slow one inside the loop and they eventually meet at the same node — that meeting is the proof of a cycle.
Why must they meet? Once both pointers are inside the cycle, the gap between them shrinks by exactly one node per step (fast gains one on slow each step). A gap that decreases by one every step inside a finite loop must hit zero — they collide.
Finding the cycle's start
This is the elegant part. Let the distance from head to the cycle entry be a, and let the meeting point be b nodes into the cycle (cycle length c).
head ──a──▶ (entry) ──b──▶ (meet)
▲ │
└──── c-b ────┘ (cycle length c)
When they meet, slow has traveled a + b and fast has traveled twice that, 2(a + b). Fast's extra distance is some whole number of laps: 2(a + b) − (a + b) = a + b = k·c. Therefore a = k·c − b, meaning the distance from the head to the entry equals the distance from the meeting point to the entry (modulo full laps).
So: reset one pointer to the head, keep the other at the meeting point, and advance both one step at a time. They meet exactly at the cycle's entry node.