Interview question: How to detect a loop in a linked list?

Question

Say you have a linked list structure in Java. It's made up of Nodes:

class Node {
    Node next;
    // some user data
}

and each Node points to the next node, except for the last Node, which has null for next. Say there is a possibility that the list can contain a loop - i.e. the final Node, instead of having a null, has a reference to one of the nodes in the list which came before it.

What's the best way of writing

boolean hasLoop(Node first)

which would return true if the given Node is the first of a list with a loop, and false otherwise? How could you write so that it takes a constant amount of space and a reasonable amount of time? (should be O(n) time)

Here's a picture of what a list with a loop looks like:

Answer 1

Take a look at Pollard's rho algorithm. It's not quite the same problem, but maybe you'll understand the logic from it, and apply it for linked lists.

(if you're lazy, you can just check out cycle detection -- check the part about the tortoise and hare.)

This only requires linear time, and 2 extra pointers.

In Java:

boolean hasLoop( Node first ) {
    if ( first == null ) return false;

    Node turtle = first;
    Node hare = first;

    while ( hare.next != null && hare.next.next != null ) {
         turtle = turtle.next;
         hare = hare.next.next;

         if ( turtle == hare ) return true;
    }

    return false;
}

(Most of the solution do not check for both next and next.next for nulls. Also, since the turtle is always behind, you don't have to check it for null -- the hare did that already.)

Answer 2

The following may not be the best method--it is O(n^2). However, it should serve to get the job done (eventually).

count_of_elements_so_far = 0;
for (each element in linked list)
{
    search for current element in first <count_of_elements_so_far>
    if found, then you have a loop
    else,count_of_elements_so_far++;
}

Answer 3

you should somehow mark nodes walked over
if next node has the mark - you have found a loop

Answer 4

Let firstnode be given.

f1 = firstnode;
f2 = firstnode.next;
while (f1 != f2 && f1.next!=null && f2.next!=null){
 f1 = f1.next;
 f2 = f2.next.next;
}
if (f1 == f2 && f1!=null && f2!=null) return "LOOP";
return "No loop";

Answer 5

You can make use of Floyd's cycle-finding algorithm, also know as tortoise and hare algorithm.

The idea is to have two references to the list and move them at different speeds. Move one forward by 1 node and the other by 2 nodes.

• If the linked list has a loop they will definitely meet.
• Else either of the two references(or their next) will become null.

Java function implementing the algorithm:

boolean hasLoop(Node first) {

    if(first == null) // list does not exist..so no loop either.
        return false;

    Node slow, fast; // create two references.

    slow = fast = first; // make both refer to the start of the list.

    while(true) {

        slow = slow.next;          // 1 hop.

        if(fast.next != null)
            fast = fast.next.next; // 2 hops.
        else
            return false;          // next node null => no loop.

        if(slow == null || fast == null) // if either hits null..no loop.
            return false;

        if(slow == fast) // if the two ever meet...we must have a loop.
            return true;
    }
}

Answer 6

Turtle and Rabbit algorithm will solve it.

Answer 7

I cannot see any way of making this take a fixed amount of time or space, both will increase with the size of the list.

I would make use of an IdentityHashMap (given that there is not yet an IdentityHashSet) and store each Node into the map. Before a node is stored you would call containsKey on it. If the Node already exists you have a cycle.

ItentityHashMap uses == instead of .equals so that you are checking where the object is in memory rather than if it has the same contents.

Answer 8

An alternative solution to the Turtle and Rabbit, not quite as nice, as I temporarily change the list:

The idea is to walk the list, and reverse it as you go. Then, when you first reach a node that has already been visited, its next pointer will point "backwards", causing the iteration to proceed towards first again, where it terminates.

Node prev = null;
Node cur = first;
while (cur != null) {
    Node next = cur.next;
    cur.next = prev;
    prev = cur;
    cur = next;
}
boolean hasCycle = prev == first && first != null && first.next != null;

// reconstruct the list
cur = prev;
prev = null;
while (cur != null) {
    Node next = cur.next;
    cur.next = prev;
    prev = cur;
    cur = next;
}

return hasCycle;

Test code:

static void assertSameOrder(Node[] nodes) {
    for (int i = 0; i < nodes.length - 1; i++) {
        assert nodes[i].next == nodes[i + 1];
    }
}

public static void main(String[] args) {
    Node[] nodes = new Node[100];
    for (int i = 0; i < nodes.length; i++) {
        nodes[i] = new Node();
    }
    for (int i = 0; i < nodes.length - 1; i++) {
        nodes[i].next = nodes[i + 1];
    }
    Node first = nodes[0];
    Node max = nodes[nodes.length - 1];

    max.next = null;
    assert !hasCycle(first);
    assertSameOrder(nodes);
    max.next = first;
    assert hasCycle(first);
    assertSameOrder(nodes);
    max.next = max;
    assert hasCycle(first);
    assertSameOrder(nodes);
    max.next = nodes[50];
    assert hasCycle(first);
    assertSameOrder(nodes);
}

Answer 9

public boolean hasLoop(Node start){   
   TreeSet<Node> set = new TreeSet<Node>();
   Node lookingAt = start;

   while (lookingAt.peek() != null){
       lookingAt = lookingAt.next;

       if (set.contains(lookingAt){
           return false;
        } else {
        set.put(lookingAt);
        }

        return true;
}   
// Inside our Node class:        
public Node peek(){
   return this.next;
}

Forgive me my ignorance (I'm still fairly new to Java and programming), but why wouldn't the above work?

I guess this doesn't solve the constant space issue... but it does at least get there in a reasonable time, correct? It will only take the space of the linked list plus the space of a set with n elements (where n is the number of elements in the linked list, or the number of elements until it reaches a loop). And for time, worst-case analysis, I think, would suggest O(nlog(n)). SortedSet look-ups for contains() are log(n) (check the javadoc, but I'm pretty sure TreeSet's underlying structure is TreeMap, whose in turn is a red-black tree), and in the worst case (no loops, or loop at very end), it will have to do n look-ups.

Answer 10

If we're allowed to embed the class Node, I would solve the problem as I've implemented it below. hasLoop() runs in O(n) time, and takes only the space of counter. Does this seem like an appropriate solution? Or is there a way to do it without embedding Node? (Obviously, in a real implementation there would be more methods, like RemoveNode(Node n), etc.)

public class LinkedNodeList {
    Node first;
    Int count;

    LinkedNodeList(){
        first = null;
        count = 0;
    }

    LinkedNodeList(Node n){
        if (n.next != null){
            throw new error("must start with single node!");
        } else {
            first = n;
            count = 1;
        }
    }

    public void addNode(Node n){
        Node lookingAt = first;

        while(lookingAt.next != null){
            lookingAt = lookingAt.next;
        }

        lookingAt.next = n;
        count++;
    }

    public boolean hasLoop(){

        int counter = 0;
        Node lookingAt = first;

        while(lookingAt.next != null){
            counter++;
            if (count < counter){
                return false;
            } else {
               lookingAt = lookingAt.next;
            }
        }

        return true;

    }



    private class Node{
        Node next;
        ....
    }

}

Answer 11

Brent's Cycle Detection Algorithm

Answer 12

The user unicornaddict has a nice algorithm above, but unfortunately it contains a bug for non-loopy lists of odd length >= 3. The problem is that fast can get "stuck" just before the end of the list, slow catches up to it, and a loop is (wrongly) detected.

Here's the corrected algorithm.

static boolean hasLoop(Node first) {

    if(first == null) // list does not exist..so no loop either.
        return false;

    Node slow, fast; // create two references.

    slow = fast = first; // make both refer to the start of the list.

    while(true) {
        slow = slow.next;          // 1 hop.
        if(fast.next == null)
            fast = null;
        else
            fast = fast.next.next; // 2 hops.

        if(fast == null) // if fast hits null..no loop.
            return false;

        if(slow == fast) // if the two ever meet...we must have a loop.
            return true;
    }
}

Answer 13

You can also look the Nivasch's algorithm here: Nivasch's algorithm.

Or you can check Gabriel Nivasch's personal homepage at The stack algorithm for cycle detection which also contains a C implementation of the algorithm.

Answer 14

Here's a refinement of the Fast/Slow solution, which correctly handles odd length lists and improves clarity.

boolean hasLoop(Node first) {
    Node slow = first;
    Node fast = first;

    while(fast != null && fast.next != null) {
        slow = slow.next;          // 1 hop
        fast = fast.next.next;     // 2 hops 

        if(slow == fast)  // fast caught up to slow, so there is a loop
            return true;
    }
    return false;  // fast reached null, so the list terminates
}

Answer 15

i think it can be done in one of the simplest way by O(n) complexity.

as you traverse the list starting from head, create a sorted list of adresses. when you insert a new adress, just check it is already there in the sorted list. the sorting is of O(logN) complexity only :-)