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

Answer 1

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 2

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 3

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 4

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)

Answer 5

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 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;

}