# Detect and Remove Loop in a Linked List

• Difficulty Level : Medium
• Last Updated : 26 Jul, 2022

Write a function detectAndRemoveLoop() that checks whether a given Linked List contains a loop and if the loop is present then removes the loop and returns true. If the list doesn’t contain a loop then it returns false. The below diagram shows a linked list with a loop. detectAndRemoveLoop() must change the below list to 1->2->3->4->5->NULL.

We also recommend reading the following post as a prerequisite to the solution discussed here.
Write a C function to detect a loop in a linked list
Before trying to remove the loop, we must detect it. Techniques discussed in the above post can be used to detect loops. To remove the loop, all we need to do is to get a pointer to the last node of the loop. For example, a node with value 5 in the above diagram. Once we have a pointer to the last node, we can make the next of this node NULL and the loop is gone.
We can easily use Hashing or Visited node techniques (discussed in the above-mentioned post) to get the pointer to the last node. The idea is simple: the very first node whose next is already visited (or hashed) is the last node.
We can also use the Floyd Cycle Detection algorithm to detect and remove the loop. In Floyd’s algo, the slow and fast pointers meet at a loop node. We can use this loop node to remove the cycle. There are following two different ways of removing the loop when Floydâ€™s algorithm is used for loop detection.

Method 1 (Check one by one) We know that Floydâ€™s Cycle detection algorithm terminates when fast and slow pointers meet at a common point. We also know that this common point is one of the loop nodes (2 or 3 or 4 or 5 in the above diagram). Store the address of this in a pointer variable say ptr2. After that start from the head of the Linked List and check for nodes one by one if they are reachable from ptr2. Whenever we find a node that is reachable, we know that this node is the starting node of the loop in the Linked List and we can get the pointer to the previous of this node.

Output:

50 20 15 4 10

Method 2 (Better Solution)

1. This method is also dependent on Floyd’s Cycle detection algorithm.
2. Detect Loop using Floyd’s Cycle detection algorithm and get the pointer to a loop node.
3. Count the number of nodes in the loop. Let the count be k.
4. Fix one pointer to the head and another to a kth node from the head.
5. Move both pointers at the same pace, they will meet at the loop starting node.
6. Get a pointer to the last node of the loop and make the next of it NULL.

Thanks to WgpShashank for suggesting this method.
Below image is a dry run of the ‘remove loop’ function in the code :

Below is the implementation of the above approach:

## C++

 #include using namespace std;  /* Link list node */struct Node {    int data;    struct Node* next;};  /* Function to remove loop. */void removeLoop(struct Node*, struct Node*);  /* This function detects and removes loop in the list  If loop was there in the list then it returns 1,  otherwise returns 0 */int detectAndRemoveLoop(struct Node* list){    struct Node *slow_p = list, *fast_p = list;      // Iterate and find if loop exists or not    while (slow_p && fast_p && fast_p->next) {        slow_p = slow_p->next;        fast_p = fast_p->next->next;          /* If slow_p and fast_p meet at some point then there           is a loop */        if (slow_p == fast_p) {            removeLoop(slow_p, list);              /* Return 1 to indicate that loop is found */            return 1;        }    }      /* Return 0 to indicate that there is no loop*/    return 0;}  /* Function to remove loop. loop_node --> Pointer to one of the loop nodes head -->  Pointer to the start node of the linked list */void removeLoop(struct Node* loop_node, struct Node* head){    struct Node* ptr1 = loop_node;    struct Node* ptr2 = loop_node;      // Count the number of nodes in loop    unsigned int k = 1, i;    while (ptr1->next != ptr2) {        ptr1 = ptr1->next;        k++;    }      // Fix one pointer to head    ptr1 = head;      // And the other pointer to k nodes after head    ptr2 = head;    for (i = 0; i < k; i++)        ptr2 = ptr2->next;      /*  Move both pointers at the same pace,      they will meet at loop starting node */    while (ptr2 != ptr1) {        ptr1 = ptr1->next;        ptr2 = ptr2->next;    }      // Get pointer to the last node    while (ptr2->next != ptr1)        ptr2 = ptr2->next;      /* Set the next node of the loop ending node      to fix the loop */    ptr2->next = NULL;}  /* Function to print linked list */void printList(struct Node* node){    // Print the list after loop removal    while (node != NULL) {        cout << node->data << " ";        node = node->next;    }}  struct Node* newNode(int key){    struct Node* temp = new Node();    temp->data = key;    temp->next = NULL;    return temp;}  // Driver Codeint main(){    struct Node* head = newNode(50);    head->next = newNode(20);    head->next->next = newNode(15);    head->next->next->next = newNode(4);    head->next->next->next->next = newNode(10);      /* Create a loop for testing */    head->next->next->next->next->next = head->next->next;      detectAndRemoveLoop(head);      cout << "Linked List after removing loop \n";    printList(head);    return 0;}  // This code has been contributed by Striver

## Python3

 # Python program to detect and remove loop in linked list  # Node class class Node:      # Constructor to initialize the node object    def __init__(self, data):        self.data = data        self.next = None  class LinkedList:      # Function to initialize head    def __init__(self):        self.head = None      def detectAndRemoveLoop(self):        slow_p = fast_p = self.head                  while(slow_p and fast_p and fast_p.next):            slow_p = slow_p.next            fast_p = fast_p.next.next              # If slow_p and fast_p meet at some point then            # there is a loop            if slow_p == fast_p:                self.removeLoop(slow_p)                          # Return 1 to indicate that loop is found                return 1                  # Return 0 to indicate that there is no loop        return 0       # Function to remove loop    # loop_node --> pointer to one of the loop nodes    # head --> Pointer to the start node of the linked list    def removeLoop(self, loop_node):        ptr1 = loop_node        ptr2 = loop_node                  # Count the number of nodes in loop        k = 1         while(ptr1.next != ptr2):            ptr1 = ptr1.next            k += 1          # Fix one pointer to head        ptr1 = self.head                  # And the other pointer to k nodes after head        ptr2 = self.head        for i in range(k):            ptr2 = ptr2.next          # Move both pointers at the same place        # they will meet at loop starting node        while(ptr2 != ptr1):            ptr1 = ptr1.next            ptr2 = ptr2.next          # Get pointer to the last node        while(ptr2.next != ptr1):            ptr2 = ptr2.next          # Set the next node of the loop ending node        # to fix the loop        ptr2.next = None      # Function to insert a new node at the beginning    def push(self, new_data):        new_node = Node(new_data)        new_node.next = self.head        self.head = new_node      # Utility function to print the LinkedList    def printList(self):        temp = self.head        while(temp):            print(temp.data, end = ' ')            temp = temp.next    # Driver programllist = LinkedList()llist.push(10)llist.push(4)llist.push(15)llist.push(20)llist.push(50)  # Create a loop for testingllist.head.next.next.next.next.next = llist.head.next.next  llist.detectAndRemoveLoop()  print("Linked List after removing loop")llist.printList()  # This code is contributed by Nikhil Kumar Singh(nickzuck_007)

## Javascript



Output

50 20 15 4 10

Method 3 (Optimized Method 2: Without Counting Nodes in Loop)
We do not need to count the number of nodes in Loop. After detecting the loop, if we start the slow pointer from the head and move both slow and fast pointers at the same speed until fast don’t meet, they would meet at the beginning of the loop.

How does this work?
Let slow and fast meet at some point after Floyd’s Cycle finding algorithm. The below diagram shows the situation when the cycle is found.

Situation when cycle is found

We can conclude below from the above diagram

Distance traveled by fast pointer = 2 * (Distance traveled
by slow pointer)

(m + n*x + k) = 2*(m + n*y + k)

Note that before meeting the point shown above, fast
was moving at twice speed.

x -->  Number of complete cyclic rounds made by
fast pointer before they meet first time

y -->  Number of complete cyclic rounds made by
slow pointer before they meet first time

From the above equation, we can conclude below

m + k = (x-2y)*n

Which means m+k is a multiple of n.
Thus we can write, m + k = i*n or m = i*n - k.
Hence, distance moved by slow pointer: m, is equal to distance moved by fast pointer:
i*n - k or (i-1)*n + n - k (cover the loop completely i-1 times and start from n-k).

So if we start moving both pointers again at same speed such that one pointer (say slow) begins from head node of linked list and other pointer (say fast) begins from meeting point. When the slow pointer reaches the beginning of the loop (has made m steps), the fast pointer would have made also moved m steps as they are now moving at the same pace. Since m+k is a multiple of n and fast starts from k, they would meet at the beginning. Can they meet before also? No because slow pointer enters the cycle first time after m steps.

## Java

 // Java program to detect // and remove loop in linked list  class LinkedList {      static Node head;      static class Node {          int data;        Node next;          Node(int d)        {            data = d;            next = null;        }    }      // Function that detects loop in the list    void detectAndRemoveLoop(Node node)    {          // If list is empty or has only one node        // without loop        if (node == null || node.next == null)            return;          Node slow = node, fast = node;          // Move slow and fast 1 and 2 steps        // ahead respectively.        slow = slow.next;        fast = fast.next.next;          // Search for loop using slow and fast pointers        while (fast != null && fast.next != null) {            if (slow == fast)                break;              slow = slow.next;            fast = fast.next.next;        }           /* If loop exists */        if (slow == fast) {            slow = node;            if (slow != fast) {                while (slow.next != fast.next) {                    slow = slow.next;                    fast = fast.next;                }                /* since fast->next is the looping point */                fast.next = null; /* remove loop */            }              /* This case is added if fast and slow pointer meet at first position. */            else {                while(fast.next != slow) {                    fast = fast.next;                }                fast.next = null;            }        }    }      // Function to print the linked list    void printList(Node node)    {        while (node != null) {            System.out.print(node.data + " ");            node = node.next;        }    }      // Driver code    public static void main(String[] args)    {        LinkedList list = new LinkedList();        list.head = new Node(50);        list.head.next = new Node(20);        list.head.next.next = new Node(15);        list.head.next.next.next = new Node(4);        list.head.next.next.next.next = new Node(10);          // Creating a loop for testing        head.next.next.next.next.next = head.next.next;        list.detectAndRemoveLoop(head);        System.out.println("Linked List after removing loop : ");        list.printList(head);    }}  // This code has been contributed by Mayank Jaiswal

## C#

 // C# program to detect and remove loop in linked listusing System;  public class LinkedList {      public Node head;      public class Node {          public int data;        public Node next;          public Node(int d)        {            data = d;            next = null;        }    }      // Function that detects loop in the list    void detectAndRemoveLoop(Node node)    {          // If list is empty or has only one node        // without loop        if (node == null || node.next == null)            return;          Node slow = node, fast = node;          // Move slow and fast 1 and 2 steps        // ahead respectively.        slow = slow.next;        fast = fast.next.next;          // Search for loop using slow and fast pointers        while (fast != null && fast.next != null) {            if (slow == fast)                break;              slow = slow.next;            fast = fast.next.next;        }          /* If loop exists */        if (slow == fast) {            slow = node;            while (slow.next != fast.next) {                slow = slow.next;                fast = fast.next;            }              /* since fast->next is the looping point */            fast.next = null; /* remove loop */        }    }      // Function to print the linked list    void printList(Node node)    {        while (node != null) {            Console.Write(node.data + " ");            node = node.next;        }    }      // Driver program to test above functions    public static void Main(String[] args)    {        LinkedList list = new LinkedList();        list.head = new Node(50);        list.head.next = new Node(20);        list.head.next.next = new Node(15);        list.head.next.next.next = new Node(4);        list.head.next.next.next.next = new Node(10);          // Creating a loop for testing        list.head.next.next.next.next.next = list.head.next.next;        list.detectAndRemoveLoop(list.head);        Console.WriteLine("Linked List after removing loop : ");        list.printList(list.head);    }}  // This code contributed by Rajput-Ji

## Javascript



Output

50 20 15 4 10

We can hash the addresses of the linked list nodes in an unordered map and just check if the element already exists in the map. If it exists, we have reached a node that already exists by a cycle, hence we need to make the last node’s next pointer NULL.

## Java

 // Java program to detect  and remove loop in a linked listimport java.util.*;  public class LinkedList {      static Node head; // head of list      /* Linked list Node*/    static class Node {        int data;        Node next;        Node(int d)        {            data = d;            next = null;        }    }      /* Inserts a new Node at front of the list. */    static public void push(int new_data)    {        /* 1 & 2: Allocate the Node &                 Put in the data*/        Node new_node = new Node(new_data);          /* 3. Make next of new Node as head */        new_node.next = head;          /* 4. Move the head to point to new Node */        head = new_node;    }      // Function to print the linked list    void printList(Node node)    {        while (node != null) {            System.out.print(node.data + " ");            node = node.next;        }    }      // Returns true if the loop is removed from the linked    // list else returns false.    static boolean removeLoop(Node h)    {        HashSet s = new HashSet();        Node prev = null;        while (h != null) {            // If we have already has this node            // in hashmap it means their is a cycle and we            // need to remove this cycle so set the next of            // the previous pointer with null.              if (s.contains(h)) {                prev.next = null;                return true;            }              // If we are seeing the node for            // the first time, insert it in hash            else {                s.add(h);                prev = h;                h = h.next;            }        }          return false;    }      /* Driver program to test above function */    public static void main(String[] args)    {        LinkedList llist = new LinkedList();          llist.push(20);        llist.push(4);        llist.push(15);        llist.push(10);          /*Create loop for testing */        llist.head.next.next.next.next = llist.head;          if (removeLoop(head)) {            System.out.println("Linked List after removing loop");            llist.printList(head);        }        else            System.out.println("No Loop found");    }}  // This code is contributed by Animesh Nag.



Output