Finding loop in a singly linked-list

Question

How can I detect that whether a singly linked-list has loop or not?? If it has loop then how to find the point of origination of the loop i.e. the node from which the loop has started.

Answer 1

You can detect it by simply running two pointers through the list, this process is known as the tortoise and hare algorithm after the fable of the same name:

• First off, check if the list is empty (head is null). If so, no cycle exists, so stop now.
• Otherwise, start the first pointer tortoise on the first node head, and the second pointer hare on the second node head.next.
• Then loop continuously until hare is null (which may be already true in a one-element list), advancing tortoise by one and hare by two in each iteration. The hare is guaranteed to reach the end first (if there is an end) since it started ahead and runs faster.
• If there is no end (i.e., if there is a cycle), they will eventually point to the same node and you can stop, knowing you have found a node somewhere within the cycle.

Consider the following loop which starts at 3:

head -> 1 -> 2 -> 3 -> 4 -> 5
                  ^         |
                  |         V
                  8 <- 7 <- 6

Starting tortoise at 1 and hare at 2, they take on the following values:

(tortoise,hare) = (1,2) (2,4) (3,6) (4,8) (5,4) (6,6)

Because they become equal at (6,6), and since hare should always be beyond tortoise in a non-looping list, it means you've discovered a cycle.

The pseudo-code will go something like this:

def hasLoop (head):
  return false if head = null           # Empty list has no loop.

  tortoise = head                       # tortoise initially first element.
  hare = tortoise.next                  # Set hare to second element.

  while hare != null:                   # Go until hare reaches end.
    return false if hare.next = null    # Check enough left for hare move.
    hare = hare.next.next               # Move hare forward two.

    tortoise = tortoise.next            # Move tortoise forward one.

    return true if hare = tortoise      # Same means loop found.
  endwhile

  return false                          # Loop exit means no loop.
enddef

The time complexity for this algorithm is O(n) since the number of nodes visited (by tortoise and hare) is proportional to the number of nodes.

---

Once you know a node within the loop, there's also an O(n) guaranteed method to find the start of the loop.

Let's return to the original position after you've found an element somewhere in the loop but you're not sure where the start of the loop is.

head -> 1 -> 2 -> 3 -> 4 -> 5
                  ^         |
                  |         V
                  8 <- 7 <- 6
                             \
                              x (where hare and tortoise met).

This is the process to follow:

• Advance hare and set size to 1.
• Then, as long as hare and tortoise are different, continue to advance hare, increasing size each time. This eventually gives the size of the cycle, six in this case.
• At this point, if size is 1, that means you must already be at the start of the cycle (in a cycle of size one, there is only one possible node that can be in the cycle so it must be the first one). In this case, you simply return hare as the start, and skip the rest of the steps below.
• Otherwise, set both hare and tortoise to the first element of the list and advance hare exactly size times (to the 7 in this case). This gives two pointers that are different by exactly the size of the cycle.
• Then, as long as hare and tortoise are different, advance them both together (with the hare running at a more sedate pace, the same speed as the tortoise - I guess it's tired from its first run). Since they will remain exactly size elements apart from each other at all times, tortoise will reach the start of the cycle at exactly the same time as hare returns to the start of the cycle.

You can see that with the following walkthrough:

size  tortoise  hare  comment
----  --------  ----  -------
   6         1     1  initial state
                   7  advance hare by six
             2     8  1/7 different, so advance both together
             3     3  2/8 different, so advance both together
                      3/3 same, so exit loop

Hence 3 is the start point of the cycle and, since both those operations (the cycle detection and cycle start discovery) are O(n) and performed sequentially, the whole thing taken together is also O(n).

---

If you want a more formal proof that this works, you can examine the following resources:

• a question on our sister site;
• the Wikipedia cycle detection page; or
• "The Tortoise and the Hare Algorithm" by Peter Gammie, April 17, 2016.

If you're simply after support for the method (not formal proof), you can run the following Python 3 program which evaluates its workability for a large number of sizes (how many elements in the cycle) and lead-ins (elements before the cycle start).

You'll find it always finds a point where the two pointers meet:

def nextp(p, ld, sz):
    if p == ld + sz:
        return ld
    return p + 1

for size in range(1,1001):
    for lead in range(1001):
        p1 = 0
        p2 = 0
        while True:
            p1 = nextp(p1, lead, size)
            p2 = nextp(nextp(p2, lead, size), lead, size)
            if p1 == p2:
                print("sz = %d, ld = %d, found = %d" % (size, lead, p1))
                break

Answer 2

The selected answer gives an O(n*n) solution to find the start node of the cycle. Here's an O(n) solution:

Once we find the slow A and fast B meet in the cycle, make one of them still and the other continue to go one step each time, to decide the perimeter of the cycle, say, P.

Then we put a node at the head and let it go P steps, and put another node at the head. We advance these two nodes both one step each time, when they first meet, it's the start point of the cycle.

Answer 3

You can use hash map also to finding whether a link list have loop or not below function uses hash map to find out whether link list have loop or not

    static bool isListHaveALoopUsingHashMap(Link *headLink) {

        map<Link*, int> tempMap;
        Link * temp;
        temp = headLink;
        while (temp->next != NULL) {
            if (tempMap.find(temp) == tempMap.end()) {
                tempMap[temp] = 1;
            } else {
                return 0;
            }
            temp = temp->next;
        }
        return 1;
    }

Two pointer method is best approach because time complexity is O(n) Hash Map required addition O(n) space complexity.

Answer 4

I read this answer in Data structure book by Narasimha Karamanchi.

We can use Floyd cycle finding algorithm, also known as tortoise and hare algorithm. In this, two pointers are used; one (say slowPtr) is advanced by a single node, and another (say fastPtr) is advanced by two nodes. If any loop is present in the single linked list, they both will surely meet at some point.

struct Node{
int data;
struct Node *next;

}

 // program to find the begin of the loop

 int detectLoopandFindBegin(struct Node *head){
      struct Node *slowPtr = head, *fastPtr = head;
      int loopExists = 0;
      // this  while loop will find if  there exists a loop or not.
      while(slowPtr && fastPtr && fastPtr->next){                                                  
        slowPtr = slowPtr->next;                      
        fastPtr = fastPtr->next->next;
        if(slowPtr == fastPtr)
        loopExists = 1;
        break;
      }

If there exists any loop then we point one of the pointers to the head and now advance both of them by single node. The node at which they will meet will be the start node of the loop in the single linked list.

        if(loopExists){      
             slowPtr = head;
             while(slowPtr != fastPtr){
               fastPtr = fastPtr->next;
               slowPtr = slowPtr->next;
             }
             return slowPtr;
          }
         return NULL;
        }

Answer 5

For the most part all the previous answers are correct but here is a simplified version of the logic with visual & code (for Python 3.7)

The logic is very simple as others explained it. I'm gonna create Tortoise/slow and Hare/fast. If we move two pointers with different speed then eventually fast will meet the slow !! you can also think of this as two runners in a tack circular field. If the fast runner keeps going in circle then it will meet/pass the slow runner.

So, we will move Tortoise/slow pointer with speed 1 for each iteration while we keep incrementing or move the Hare/fast pointer with speed of 2. Once they meet we know there is a cycle. This is also known as Floyd's cycle-finding algorithm

Here is the Python code that does this (notice has_cycle method is the main part):

#!/usr/bin/env python3
class Node:
    def __init__(self, data = None):
        self.data = data
        self.next = None
    def strnode (self):
        print(self.data)


class LinkedList:
    def __init__(self):
        self.numnodes = 0
        self.head = None


    def insertLast(self, data):
        newnode = Node(data)
        newnode.next = None
        if self.head == None:
            self.head = newnode
            return

        lnode = self.head
        while lnode.next != None :
            lnode = lnode.next
        lnode.next = newnode # new node is now the last node
        self.numnodes += 1

    def has_cycle(self):    
        slow, fast = self.head ,self.head  
        while fast != None:       
            if fast.next != None:
                 fast = fast.next.next
            else:
                 return False
            slow = slow.next  
            if slow == fast:
                print("--slow",slow.data, "fast",fast.data) 
                return True    
        return False


linkedList = LinkedList()
linkedList.insertLast("1")
linkedList.insertLast("2")
linkedList.insertLast("3")


# Create a loop for testing 
linkedList.head.next.next.next = linkedList.head; 
#let's check and see !
print(linkedList.has_cycle())

Answer 6

Following code will find whether there is a loop in SLL and if there, will return then starting node.

int find_loop(Node *head){

    Node * slow = head;
    Node * fast =  head;
    Node * ptr1;
    Node * ptr2;
    int k =1, loop_found =0, i;

    if(!head) return -1;

    while(slow && fast && fast->next){
            slow = slow->next;
        /*Moving fast pointer two steps at a time */
            fast = fast->next->next;
            if(slow == fast){
                    loop_found = 1;
                    break;
            }

    }

    if(loop_found){
    /* We have detected a loop */
    /*Let's count the number of nodes in this loop node */

            ptr1  = fast;
            while(ptr1 && ptr1->next != slow){
                    ptr1 = ptr1->next;
                    k++;
            }
    /* Now move the other pointer by K nodes */
            ptr2 = head;

            ptr1  = head;
            for(i=0; i<k; i++){
                    ptr2 = ptr2->next;
            }

    /* Now if we move ptr1 and ptr2 with same speed they will meet at start of loop */

            while(ptr1 != ptr2){
                    ptr1  = ptr1->next;
                    ptr2 =  ptr2->next;
            }

    return ptr1->data;

}

Answer 7

boolean hasLoop(Node *head)
    {
      Node *current = head;
      Node *check = null;
      int firstPtr = 0;
      int secondPtr = 2;
      do {
        if (check == current) return true;
        if (firstPtr >= secondPtr){
            check = current;
            firstPtr = 0;
            secondPtr= 2*secondPtr;
        }
        firstPtr ++;
      } while (current = current->next());
      return false;
    }

Another O(n) solution.

Answer 8

As I viewed the selected answer, I tried a couple of examples and found that:
If (A1,B1), (A2,B2) ... (AN, BN) are the traversals of pointers A and B
where A steps 1 element and B steps 2 elements, and, Ai and Bj are the nodes traversed by A and B, and AN=BN.
Then, the node from where the loop starts is Ak, where k = floor(N/2).

Answer 9

ok - I ran into this in an interview yesterday - no reference material available and I came up with a very different answer (while driving home of course...) Since the linked lists are NORMALLY (not always I admit) allocated using malloc logic then we know that the granularity of the allocations is known. On most systems this is 8 bytes - this means that the bottom 3 bits are always zeros. Consider - if we place the linked list in a class to control access and use a mask of 0x0E ored into the next address then we can use the lower 3 bits to store a break crumb Thus we can write a method that will store our last breadcrumb - say 1 or 2 - and alternate them. Our method that checks for a loop can then step through each node (using our next method) and check if the next address contains the current breadcrumb - if it does we have a loop - if it does not then we would mask the lower 3 bits and add our current breadcrumb. The breadcrumb checking algorithm would have to be single threaded as you could not run two of them at once but it would allow other threads to access the list async - with the usual caveats about adding/deleting nodes.
What do you think? If others feel this is a valid solution I can write up the sample class ... Just think sometimes a fresh approach is good and am always willing to be told I have just missed the point... Thanks All Mark

Answer 10

Another solution

Detecting a Loop:

1. create a list
2. loop through the linkedlist and keep on adding the node to the list.
3. If the Node is already present in the List, we have a loop.

Removal of loop:

1. In the Step#2 above, while loop through the linked list we are also keep track of the previous node.

2. Once we detect the loop in Step#3, set previous node's next value to NULL

   #code

   def detect_remove_loop(head)

       cur_node = head
       node_list = []
   
       while cur_node.next is not None:
           prev_node = cur_node
           cur_node = cur_node.next
           if cur_node not in node_list:
               node_list.append(cur_node)
           else:
               print('Loop Detected')
               prev_node.next = None
               return
   
       print('No Loop detected')
   

Answer 11

Firstly, Create a Node

struct Node { 
    int data; 
    struct Node* next; 
}; 

Initialize head pointer globally

Struct Node* head = NULL;

Insert some data in Linked List

void insert(int newdata){

    Node* newNode = new Node();
    newNode->data = newdata;
    newNode->next = head;
    head = newNode;
}

Create a function detectLoop()

void detectLoop(){
    if (head == NULL || head->next == NULL){
        cout<< "\nNo Lopp Found in Linked List";
    }
    else{
        Node* slow = head;
        Node* fast = head->next;
        while((fast && fast->next) && fast != NULL){
            if(fast == slow){
                cout<<"Loop Found";
                break;
            }
            fast = fast->next->next;
            slow = slow->next;
        }
        if(fast->next == NULL){
            cout<<"Not Found";
        }
    }
}

Call the function from main()

int main() 
{ 
    insert(4);
    insert(3);
    insert(2);
    insert(1);

    //Created a Loop for Testing, Comment the next line to check the unloop linkedlist
    head->next->next->next->next = head->next;

    detectLoop();
    //If you uncomment the display function and make a loop in linked list and then run the code you will find infinite loop 
    //display();
} 

Answer 12

                bool FindLoop(struct node *head)
                {
                    struct node *current1,*current2;

                    current1=head;
                    current2=head;

                    while(current1!=NULL && current2!= NULL && current2->next!= NULL)
                    { 
                          current1=current1->next;
                          current2=current2->next->next;

                          if(current1==current2)
                          {
                                return true;
                          }
                    }

                    return false;
                }

Answer 13

A quite different method:- Reverse the linked list. While reversing if you reach the head again then there is a loop in the list, if you get NULL then there is no loop. The total time complexity is O(n)