Detect and Remove Loop in a Linked List
Write a function detectAndRemoveLoop() that checks whether a given Linked List contains loop and if loop is present then removes the loop and returns true. If the list doesn’t contain loop then it returns false. Below diagram shows a linked list with a loop. detectAndRemoveLoop() must change the below list to 1->2->3->4->5->NULL.
Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.
We also recommend to read following post as a prerequisite of the solution discussed here.
Write a C function to detect 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 loop. To remove loop, all we need to do is to get pointer to the last node of the loop. For example, node with value 5 in the above diagram. Once we have pointer to the last node, we can make the next of this node as NULL and 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. Idea is simple: the very first node whose next is already visited (or hashed) is the last node.
We can also use Floyd Cycle Detection algorithm to detect and remove the loop. In the Floyd’s algo, the slow and fast pointers meet at a loop node. We can use this loop node to remove cycle. There are following two different ways of removing loop when Floyd’s algorithm is used for Loop detection.
Method 1 (Check one by one) We know that Floyd’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 Linked List and we can get the pointer to the previous of this node.
Output:
Linked List after removing loop
50 20 15 4 10
Method 2 (Better Solution)
- This method is also dependent on Floyd’s Cycle detection algorithm.
- Detect Loop using Floyd’s Cycle detection algorithm and get the pointer to a loop node.
- Count the number of nodes in loop. Let the count be k.
- Fix one pointer to the head and another to a kth node from the head.
- Move both pointers at the same pace, they will meet at loop starting node.
- Get a pointer to the last node of the loop and make next of it as NULL.
Thanks to WgpShashank for suggesting this method.
Below image is a dry run of ‘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 Code
int 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
|
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 Code
int 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
|
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
int detectAndRemoveLoop(Node node)
{
Node slow = node, fast = node;
while (slow != null && fast != null && fast.next != null) {
slow = slow.next;
fast = fast.next.next;
// If slow and fast meet at same point then loop is present
if (slow == fast) {
removeLoop(slow, node);
return 1;
}
}
return 0;
}
// Function to remove loop
void removeLoop(Node loop, Node head)
{
Node ptr1 = loop;
Node ptr2 = loop;
// Count the number of nodes in loop
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 the linked list
void printList(Node node)
{
while (node != null) {
System.out.print(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
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
|
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 program
llist = LinkedList()
llist.push(10)
llist.push(4)
llist.push(15)
llist.push(20)
llist.push(50)
# Create a loop for testing
llist.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)
|
C#
// A C# program to detect and remove loop in linked list
using System;
public class LinkedList {
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
int detectAndRemoveLoop(Node node)
{
Node slow = node, fast = node;
while (slow != null && fast != null && fast.next != null) {
slow = slow.next;
fast = fast.next.next;
// If slow and fast meet at same
// point then loop is present
if (slow == fast) {
removeLoop(slow, node);
return 1;
}
}
return 0;
}
// Function to remove loop
void removeLoop(Node loop, Node head)
{
Node ptr1 = loop;
Node ptr2 = loop;
// Count the number of nodes in loop
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 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
// Javascript program to detect and
// remove loop in linked list
var head;
class Node
{
constructor(val)
{
this.data = val;
this.next = null;
}
}
// Function that detects loop in the list
function detectAndRemoveLoop(node)
{
var slow = node, fast = node;
while (slow != null &&
fast != null &&
fast.next != null)
{
slow = slow.next;
fast = fast.next.next;
// If slow and fast meet at same
// point then loop is present
if (slow == fast)
{
removeLoop(slow, node);
return 1;
}
}
return 0;
}
// Function to remove loop
function removeLoop(loop, head)
{
var ptr1 = loop;
var ptr2 = loop;
// Count the number of nodes in loop
var 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 the linked list
function printList(node)
{
while (node != null)
{
document.write(node.data + " ");
node = node.next;
}
}
// Driver code
head = new Node(50);
head.next = new Node(20);
head.next.next = new Node(15);
head.next.next.next = new Node(4);
head.next.next.next.next = new Node(10);
// Creating a loop for testing
head.next.next.next.next.next = head.next.next;
detectAndRemoveLoop(head);
document.write("Linked List after removing loop : ");
printList(head);
// This code is contributed by todaysgaurav
|
Output:
Linked List after removing loop
50 20 15 4 10
Method 3 (Optimized Method 2: Without Counting Nodes in Loop)
We do not need to count number of nodes in Loop. After detecting the loop, if we start slow pointer from head and move both slow and fast pointers at 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. Below diagram shows the situation when cycle is found.
We can conclude below from 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 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 slow pointer reaches beginning of loop (has made m steps), fast pointer would have made also moved m steps as they are now moving 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.
C++
// C++ program to detect and remove loop
#include
using namespace std;
struct Node {
int key;
struct Node* next;
};
Node* newNode(int key)
{
Node* temp = new Node;
temp->key = key;
temp->next = NULL;
return temp;
}
// A utility function to print a linked list
void printList(Node* head)
{
while (head != NULL) {
cout << head->key << " ";
head = head->next;
}
cout << endl;
}
// Function to detect and remove loop
// in a linked list that may contain loop
void detectAndRemoveLoop(Node* head)
{
// If list is empty or has only one node
// without loop
if (head == NULL || head->next == NULL)
return;
Node *slow = head, *fast = head;
// 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 && fast->next) {
if (slow == fast)
break;
slow = slow->next;
fast = fast->next->next;
}
/* If loop exists */
if (slow == fast)
{
slow = head;
// this check is needed when slow
// and fast both meet at the head of the LL
// eg: 1->2->3->4->5 and then
// 5->next = 1 i.e the head of the LL
if(slow == fast) {
while(fast->next != slow) fast = fast->next;
}
else {
while (slow->next != fast->next) {
slow = slow->next;
fast = fast->next;
}
}
/* since fast->next is the looping point */
fast->next = NULL; /* remove loop */
}
}
/* Driver program to test above function*/
int main()
{
Node* head = newNode(50);
head->next = head;
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;
detectAndRemoveLoop(head);
printf("Linked List after removing loop \n");
printList(head);
return 0;
}
|
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
|
Python3
# Python program to detect and remove loop
# 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
# 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
def detectAndRemoveLoop(self):
if self.head is None:
return
if self.head.next is None:
return
slow_p = self.head
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:
slow_p = self.head
# Finding the beginning of the loop
while (slow_p.next != fast_p.next):
slow_p = slow_p.next
fast_p = fast_p.next
# Sinc fast.next is the looping point
fast_p.next = None # Remove loop
# Utility function to print the LinkedList
def printList(self):
temp = self.head
while(temp):
print(temp.data, end = ' ')
temp = temp.next
# Driver program
llist = LinkedList()
llist.head = Node(50)
llist.head.next = Node(20)
llist.head.next.next = Node(15)
llist.head.next.next.next = Node(4)
llist.head.next.next.next.next = Node(10)
# Create a loop for testing
llist.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)
|
C#
// C# program to detect and remove loop in linked list
using 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
// javascript program to detect
// and remove loop in linked list
var head;
class Node
{
constructor(val)
{
this.data = val;
this.next = null;
}
}
// Function that detects loop in the list
function detectAndRemoveLoop(node) {
// If list is empty or has only one node
// without loop
if (node == null || node.next == null)
return;
var 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
function printList(node) {
while (node != null) {
document.write(node.data + " ");
node = node.next;
}
}
// Driver code
head = new Node(50);
head.next = new Node(20);
head.next.next = new Node(15);
head.next.next.next = new Node(4);
head.next.next.next.next = new Node(10);
// Creating a loop for testing
head.next.next.next.next.next = head.next.next;
detectAndRemoveLoop(head);
document.write("Linked List after removing loop : ");
printList(head);
// This code is contributed by umadevi9616
|
Output:
Linked List after removing loop
50 20 15 4 10
Method 4 Hashing: Hash the address of the linked list nodes
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 which already exists by a cycle, hence we need to make the last node’s next pointer NULL.
C++
// C++ program to detect and remove loop
#include
using namespace std;
struct Node {
int key;
struct Node* next;
};
Node* newNode(int key)
{
Node* temp = new Node;
temp->key = key;
temp->next = NULL;
return temp;
}
// A utility function to print a linked list
void printList(Node* head)
{
while (head != NULL) {
cout << head->key << " ";
head = head->next;
}
cout << endl;
}
// Function to detect and remove loop
// in a linked list that may contain loop
void hashAndRemove(Node* head)
{
// hash map to hash addresses of the linked list nodes
unordered_map node_map;
// pointer to last node
Node* last = NULL;
while (head != NULL) {
// if node not present in the map, insert it in the map
if (node_map.find(head) == node_map.end()) {
node_map[head]++;
last = head;
head = head->next;
}
// if present, it is a cycle, make the last node's next pointer NULL
else {
last->next = NULL;
break;
}
}
}
/* Driver program to test above function*/
int main()
{
Node* head = newNode(50);
head->next = head;
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;
// printList(head);
hashAndRemove(head);
printf("Linked List after removing loop \n");
printList(head);
return 0;
}
|
Java
// Java program to detect and remove loop in a linked list
import 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.
|
C#
// C# program to detect and remove loop in a linked list
using System;
using System.Collections.Generic;
public class LinkedList {
public Node head; // head of list
/* Linked list Node*/
public class Node {
public int data;
public Node next;
public Node(int d)
{
data = d;
next = null;
}
}
// Function to print the linked list
void printList(Node node)
{
while (node != null) {
Console.Write(node.data + " ");
node = node.next;
}
}
// Returns true if the loop is removed from the linked
// list else returns false.
bool 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()
{
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);
/*Create loop for testing */
list.head.next.next.next.next.next = list.head.next.next;
if (list.removeLoop(list.head)) {
Console.WriteLine("Linked List after removing loop");
list.printList(list.head);
}
else
Console.WriteLine("No Loop found");
}
}
// This code is contributed by ihritik
|
Javascript
// javascript program to detect and remove loop in a linked list class LinkedList {
/* Linked list Node */
class Node {
constructor(val) {
this.data = val;
this.next = null;
}
}
var head; // head of list
/* Inserts a new Node at front of the list. */
function push(new_data) {
/*
* 1 & 2: Allocate the Node & Put in the data
*/
var 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;
return head;
}
// Function to print the linked list
function printList(node) {
while (node != null) {
document.write(node.data + " ");
node = node.next;
}
}
// Returns true if the loop is removed from the linked
// list else returns false.
function removeLoop(h)
{
var s = new Set();
var 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.has(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 */
head = push(50);
head.next = push(20);
head.next.next=push(4);
head.next.next.next=push(15);
head.next.next.next.next=push(10);
/* Create loop for testing */
head.next.next.next.next.next = head.next.next;
if (removeLoop(head)) {
document.write("Linked List after removing loop ");
printList(head);
} else
document.write("No Loop found");
// This code is contributed by gauravrajput1
|
Output
Linked List after removing loop
50 20 15 4 10
We Thank Shubham Agrawal for suggesting this solution.
Thanks to Gaurav Ahirwar for suggesting the above solution.
Please write comments if you find the above codes/algorithms incorrect, or find other ways to solve the same problem.
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Linked List
1. Using Hashing
A simple solution is to use hashing. The idea is to traverse the given linked list and insert each encountered node into a hash set. If the node is already present in the HashSet, that means it is seen before and points to the first node in the cycle. To break the cycle, set the next pointer of the previous node to null.
Following is the implementation of the above idea in C++, Java, and Python. The code uses two pointers: curr and prev, where curr is the main pointer running down the list, and prev trails it.
C++
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 | #include #include using namespace std; // A Linked List Node struct Node { int data; Node* next; }; // Utility function to create a new node with the given data and // pushes it onto the list's front void push(Node*& headRef, int data) { // create a new linked list node from the heap Node* newNode = new Node; newNode->data = data; newNode->next = headRef; headRef = newNode; } // Utility function to print a linked list void printList(Node* head) { Node* curr = head; while (curr) { cout << curr->data << " —> "; curr = curr->next; } cout << "nullptr"; } // Function to identify and remove cycle in a linked list using hashing void removeCycle(Node* head) { Node* prev = nullptr; // previous pointer Node* curr = head;// main pointer // maintain a set to store visited nodes unordered_set<Node*> set; // traverse the list while (curr) { // set the previous pointer to null if the current node is seen before if (set.find(curr) != set.end()) { prev->next = nullptr; return; } // insert the current node into the set set.insert(curr); // update the previous pointer to the current node and // move the main pointer to the next node prev = curr; curr = curr->next; } } int main() { // total number of nodes in the linked list int n = 5; // construct a linked list Node* head = nullptr; for (int i = n; i > 0; i--) { push(head, i); } // insert cycle head->next->next->next->next->next = head->next; removeCycle(head); printList(head); return 0; } |
DownloadRun Code
Output:
1 —> 2 —> 3 —> 4 —> 5 —> nullptr