computer scientists. A singly linked list is made of nodes where each

node has a pointer to the next node (or null to end the list). A singly

linked list is often used as a stack (or last in first out queue (LIFO))

because adding a new first element, removing the existing first

element, and examining the first element are very fast O(1) operations.

When working with singly linked list, you are typically given a link

to the first node. Common operations on a singly linked list are

iterating through all the nodes, adding to the list, or deleting from

the list. Algorithms for these operations generally require a well

formed linked list. That is a linked list without loops or cycles in

it.

If a linked list has a cycle:

- The malformed linked list has no end (no node ever has a null

next_node pointer) - The malformed linked list contains two links to some node
- Iterating through the malformed linked list will yield all nodes in

the loop multiple times

A malformed linked list with a loop causes iteration over the list to

fail because the iteration will never reach the end of the list.

Therefore, it is desirable to be able to detect that a linked list is

malformed before trying an iteration. This article is a discussion of

various algorithms to detect a loop in a singly linked list.

## Incorrect "Solutions"

### Traverse the List Until the End

Just look at the entire list to see if it has and end. When it ends,

return.

// Incorrect 'solution'

function boolean hasLoop(Node startNode){

Node currentNode = startNode;

while (currentNode = currentNode.next());

return false;

}

The problem with this solution is that if the linked list does have a

loop, the program will never terminate. There is no way for this

algorithm to return true when the linked list does have a loop.

### Mark Each Node

Traverse the list and mark each node as having been seen. If you

come to a node that has already been marked, then you know that the list

has a loop.

// Incorrect 'solution'<br />function boolean hasLoop(Node startNode){<br /> Node currentNode = startNode;<br /> do {<br /> if (currentNode.seen) return true;<br /> currentNode.seen = true;<br /> } while (currentNode = currentNode.next());<br /> return false;<br />}

The problem with this solution is ensuring that "seen" is marked as

false for all the nodes before you start. If the linked list has a

loop, it isn't possible to iterate over each node to set "seen" to

"false" as an initial value for each node. It might be possible to

overcome some of this by using a big integer rather than a boolean and

using a random integer as your marker. In that case there is is a good

chance that no node will have your inital value, but a small chance that

one would and your algorithm would fail.

Even if you are able to solve the initial value problem, each node in

a linked list may not have a field to use for this purpose. Requiring

such a field in each node would mean that this is not a generic

solution. As we will see later, this field is not needed for a

perfectly correct and efficient solution anyway.

### Detect Only Full Loops

When asked to come up with a solution, a common pitfall is not

detecting all loops, but just a loop where the last node links to the

first. A loop could still occur (and not be detected) if the last

element linked to (for example) the second element.

// Incorrect 'solution'<br />function boolean hasLoop(Node startNode){<br /> Node currentNode = startNode;<br /> while (currentNode = currentNode.next()){<br /> if (currentNode == startNode) return true;<br /> }<br /> return false;<br />}

## Inefficient Solutions

### Keep a hash set of all nodes seen so far

O(n) time complexity, O(n) space complexity

Keeping a set of all the nodes have seen so far and testing to see if

the next node is in that set would be a perfectly correct solution. It

would run fast as well. However it would use enough extra space to

make a copy of the linked list. Allocating that much memory is

prohibitively expensive for large lists.

// Inefficient solution<br />function boolean hasLoop(Node startNode){<br /> HashSet nodesSeen = new HashSet();<br /> Node currentNode = startNode;<br /> do {<br /> if (nodesSeen.contains(currentNode)) return true;<br /> nodesSeen.add(currentNode);<br /> } while (currentNode = currentNode.next());<br /> return false;<br />}

### Use a doubly linked list

O(n) time complexity

Doubly linked lists make it easy to tell if there is a loop. If you

encounter any node that doesn't link to the last node you visited, you

know that there are two nodes linking to that node. Because the back

links could be initially messed up in some other way, this algorithm is

only correct if you can trust the back links. Otherwise it is just a

malformed doubly linked list finder. The singly linked list can even be

converted into a doubly linked list with little additional work. Again

this will require that we change the structure of the Node to

accomodate a second link. Something that may not be possible in all

cases. Usually a singly linked list is used because the amount of space

to allocate for each node is at a premium.

// Inefficient solution<br />function boolean hasLoop(Node startNode){<br /> Node currentNode = startNode;<br /> Node previousNode = null;<br /> do {<br /> if (previousNode && currentNode.prev() && previousNode != currentNode.prev()) return true;<br /> if (!currentNode.prev()) currentNode.setPrev(previousNode);<br /> previousNode = currentNode;<br /> } while (currentNode = currentNode.next());<br /> return false;<br />}

### Check the Entire List So Far

O(n^2) time complexity

For each node, assume that the portion of the list examined so for

has no loops and check to see if the next node creates a loop by

iterating again over the entire list up to that point.

// Inefficient solution<br />function boolean hasLoop(Node startNode){<br /> Node currentNode = startNode.next();<br /> int i=0;<br /> do {<br /> Node checkNode = startNode;<br /> int j=0;<br /> do {<br /> if (checkNode == currentNode) return true;<br /> j++;<br /> } while (j<i && checkNode = currentNode.next());<br /> i++; <br /> } while (currentNode = currentNode.next());<br /> return false;<br />}

### Reverse the list

O(n) time complexity

If you reverse the list, and remember the inital node, you will know

that there is a cycle if you get back to the first node. While

efficient, this solution changes the list. Reversing the list twice

would put the list back in its initial state, however this solution is

not appropriate for multi-threaded applications. In some cases there

may not be a way to modify nodes. Since changing the nodes is not needed

to get the answer, this solution is not recommended.

// Solution modifies the list<br />function boolean hasLoop(Node startNode){<br /> Node previousNode = null;<br /> Node currentNode = startNode;<br /> Node nextNode;<br /> if (!currentNode.next()) return false;<br /> while(currentNode){<br /> nextNode = currentNode.next();<br /> currentNode.setNext(previousNode);<br /> previousNode = currentNode;<br /> currentNode = nextNode;<br /> }<br /> return (previousNode == startNode);<br />}

Credit for this solution goes to Piyush Srivastava.

### Use Memory Allocation Information

O(n) time complexity in the amount of memory on the computer

Some programming languages allow you to see meta information about

each node -- the memory address at which it is allocated. Because each

node has a unique numeric address, it is possible to use this

information to detect cycles. For this algorithm, keep track of the

minimum memory address seen, the maximum memory address seen, and the

number of nodes seen. If more nodes have been seen than can fit in the

address space then some node must have been seen twice and there is a

cycle.

// Depends on size of available computer memory rather than size of list<br />function boolean hasLoop(Node startNode){<br /> Node currentNode = startNode;<br /> int minAddress, int maxAddress = ¤tNode;<br /> int nodesSeen = 0;<br /> while(currentNode = currentNode.next()){<br /> nodesSeen++;<br /> if (¤tNode < minAddress) minAddress = ¤tNode;<br /> if (¤tNode > maxAddress) maxAddress = ¤tNode;<br /> if (maxAddress - minAddress < nodesSeen) return true;<br /> }<br /> return false;<br />}

This algorithm relies on being able to see memomory address

information. This is not possible to implement in some programming

languages such as Java that do not make this information available. It

is likely that the entire list will be allocated close together in

memory. In such a case the implementation will run close to the running

time of the length of the list. However, if the nodes in the list are

allocated over a large memory space, the runtime of this algorithm could

be much greater than some of the best solutions.

## Best Solutions

### Catch Larger and Larger Loops

O(n) time complexity

Always store some node to check. Occasionally reset this node to

avoid the "Detect Only Full Loops" problem. When resetting it, double

the amount of time before resetting it again.

// Good solution<br />function boolean hasLoop(Node startNode){<br /> Node currentNode = startNode;<br /> Node checkNode = null;<br /> int since = 0;<br /> int sinceScale = 2;<br /> do {<br /> if (checkNode == currentNode) return true;<br /> if (since >= sinceScale){<br /> checkNode = currentNode;<br /> since = 0;<br /> sinceScale = 2*sinceScale;<br /> }<br /> since++;<br /> } while (currentNode = currentNode.next());<br /> return false;<br />}

This solution is O(n) because sinceScale grows linearly with the

number of calls to next(). Once sinceScale is greater than the size of

the loop, another n calls to next() may be required to detect the loop.

This solution requires up to 3 traversals of the list.

This solution was devised by Stephen Ostermiller and proven O(n) by

Daniel Martin.

### Catch Loops in Two Passes

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.

// Best solution<br />function boolean hasLoop(Node startNode){<br /> Node slowNode = Node fastNode1 = Node fastNode2 = startNode;<br /> while (slowNode && fastNode1 = fastNode2.next() && fastNode2 = fastNode1.next()){<br /> if (slowNode == fastNode1 || slowNode == fastNode2) return true;<br /> slowNode = slowNode.next();<br /> }<br /> return false;<br />}

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