I am sure you must know answer to this by now. Nevertheless, its very easy to miss the most valid answers to this which is the lesser known “Tortoise and Hare Algorithm”
Solution with – O(n) time complexity
Simultaneously go through the list by ones (slow iterator) and by twos (fast iterator). If there is a loop the fast iterator will go around that loop twice as fast as the slow iterator. The fast iterator will lap the slow iterator within a single pass through the cycle. Detecting a loop is then just detecting that the slow iterator has been lapped by the fast iterator.
This solution is "Floyd’s Cycle-Finding Algorithm" as published in "Non-deterministic Algorithms" by Robert W. Floyd in 1967. It is also called "The Tortoise and the Hare Algorithm".
function boolean hasLoop(Node startNode){
Node slowNode = Node fastNode1 = Node fastNode2 = startNode;
while (slowNode && fastNode1 = fastNode2.next() && fastNode2 = fastNode1.next()){
if (slowNode == fastNode1 || slowNode == fastNode2) return true;
slowNode = slowNode.next();
}
return false;
}
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