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

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())