Detect if a Linked List is Circular in DSA C++ - Build from Scratch

Detect if a Linked List is Circular

📖 Scenario: Imagine you have a chain of connected train cars. Normally, the last car is not connected back to the first car. But sometimes, the last car might connect back to an earlier car, making a loop. This is called a circular chain.In programming, a linked list is like this chain of train cars. We want to check if the linked list forms a loop (circular) or not.

🎯 Goal: You will build a simple linked list and write code to check if it is circular (has a loop) or not.

📋 What You'll Learn

Create a linked list with exactly 5 nodes with values 10, 20, 30, 40, 50

Create a boolean variable called isCircular to store the result

Write a function called checkCircular that returns true if the linked list is circular, false otherwise

Print "Circular" if the list is circular, else print "Not Circular"

💡 Why This Matters

🌍 Real World

Detecting loops in linked lists is important in many software systems to avoid infinite loops and crashes, such as in network routing, memory management, and data processing.

💼 Career

Understanding how to detect circular linked lists is a common interview question and a fundamental skill for software developers working with data structures.

1

Create a linked list with 5 nodes

Create a linked list with 5 nodes having values 10, 20, 30, 40, and 50. Use a struct called Node with an integer data and a pointer next. Create a pointer head pointing to the first node.

DSA C++

// Define struct Node
// Create nodes with values 10, 20, 30, 40, 50
// Link nodes in order
// Set head to first node
// Your code here

Hint

Think of each node as a train car. Create each car and connect it to the next.

2

Add a boolean variable to store circular check result

Create a boolean variable called isCircular and set it to false initially. This will store if the linked list is circular or not.

DSA C++

#include 
using namespace std;

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

int main() {
    Node* head = new Node{10, nullptr};
    Node* second = new Node{20, nullptr};
    Node* third = new Node{30, nullptr};
    Node* fourth = new Node{40, nullptr};
    Node* fifth = new Node{50, nullptr};

    head->next = second;
    second->next = third;
    third->next = fourth;
    fourth->next = fifth;
    fifth->next = nullptr;

    // Create boolean variable isCircular and set to false
    // Your code here

    return 0;
}

Hint

Use bool isCircular = false; to create the variable.

3

Write a function to check if the linked list is circular

Write a function called checkCircular that takes a pointer head to the linked list and returns true if the list is circular, otherwise false. Use two pointers called slow and fast. Move slow by one step and fast by two steps. If they meet, the list is circular. If fast or fast->next becomes nullptr, the list is not circular.

DSA C++

#include 
using namespace std;

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

bool checkCircular(Node* head) {
    // Use two pointers slow and fast
    // Move slow by one and fast by two
    // If they meet, return true
    // If fast or fast->next is nullptr, return false
    // Your code here
}

int main() {
    Node* head = new Node{10, nullptr};
    Node* second = new Node{20, nullptr};
    Node* third = new Node{30, nullptr};
    Node* fourth = new Node{40, nullptr};
    Node* fifth = new Node{50, nullptr};

    head->next = second;
    second->next = third;
    third->next = fourth;
    fourth->next = fifth;
    fifth->next = nullptr;

    bool isCircular = false;

    return 0;
}

Hint

Use Floyd's cycle detection method with two pointers moving at different speeds.

4

Call the function and print if the list is circular

Call the function checkCircular with head and store the result in isCircular. Then print "Circular" if isCircular is true, otherwise print "Not Circular".

DSA C++

#include 
using namespace std;

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

bool checkCircular(Node* head) {
    if (head == nullptr) return false;
    Node* slow = head;
    Node* fast = head;
    while (fast != nullptr && fast->next != nullptr) {
        slow = slow->next;
        fast = fast->next->next;
        if (slow == fast) {
            return true;
        }
    }
    return false;
}

int main() {
    Node* head = new Node{10, nullptr};
    Node* second = new Node{20, nullptr};
    Node* third = new Node{30, nullptr};
    Node* fourth = new Node{40, nullptr};
    Node* fifth = new Node{50, nullptr};

    head->next = second;
    second->next = third;
    third->next = fourth;
    fourth->next = fifth;
    fifth->next = nullptr;

    bool isCircular = false;

    // Call checkCircular and print result
    // Your code here

    return 0;
}

Hint

Call isCircular = checkCircular(head); then use if to print the correct message.

Practice

1. What does it mean if a linked list is circular?

easy

A. The linked list nodes are sorted in ascending order

B. The linked list has no nodes

C. The last node points back to one of the previous nodes, forming a loop

D. The linked list has only one node

Solution

1. Step 1: Understand linked list structure

   A linked list is a chain of nodes where each node points to the next node.

2. Step 2: Define circular linked list

   If the last node points back to any previous node, it creates a loop, making the list circular.

3. Final Answer:

   The last node points back to one of the previous nodes, forming a loop -> Option C

4. Quick Check:

   Circular linked list = last node points back [OK]

Hint: Circular means last node links back to earlier node [OK]

Common Mistakes:

• Thinking circular means sorted list
• Confusing empty list with circular list
• Assuming single node list is always circular

2. Which of the following C++ code snippets correctly checks if a linked list is circular using a slow and fast pointer approach?

easy

A. while(slow != nullptr) { slow = slow->next; if(slow == fast) return true; } return false;

B. while(fast != nullptr && fast->next != nullptr) { slow = slow->next; fast = fast->next->next; if(slow == fast) return true; } return false;

C. for(int i=0; i<10; i++) { if(slow == fast) return true; slow = slow->next; fast = fast->next; } return false;

D. if(head == nullptr) return true; else return false;

Solution

1. Step 1: Understand slow and fast pointer method

   Two pointers move at different speeds; if they meet, a loop exists.

2. Step 2: Analyze code correctness

   while(fast != nullptr && fast->next != nullptr) { slow = slow->next; fast = fast->next->next; if(slow == fast) return true; } return false; correctly moves slow by one and fast by two nodes, checking for equality inside the loop.

3. Final Answer:

   while(fast != nullptr && fast->next != nullptr) { slow = slow->next; fast = fast->next->next; if(slow == fast) return true; } return false; -> Option B

4. Quick Check:

   Slow-fast pointer meeting means circular [OK]

Hint: Fast pointer moves twice as fast; meeting means loop [OK]

Common Mistakes:

• Not checking fast and fast->next for nullptr
• Moving pointers at same speed
• Returning true without pointer meeting

3. What will be the output of the following C++ code snippet?

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

Node* head = new Node{1, nullptr};
Node* second = new Node{2, nullptr};
Node* third = new Node{3, nullptr};
head->next = second;
second->next = third;
third->next = second; // creates a loop

bool isCircular(Node* head) {
  Node* slow = head;
  Node* fast = head;
  while(fast != nullptr && fast->next != nullptr) {
    slow = slow->next;
    fast = fast->next->next;
    if(slow == fast) return true;
  }
  return false;
}

std::cout << (isCircular(head) ? "Circular" : "Not Circular") << std::endl;

medium

A. Circular

B. Not Circular

C. Compilation Error

D. Runtime Error

Solution

1. Step 1: Analyze linked list structure

   third->next points to second, creating a loop between second and third nodes.

2. Step 2: Trace slow and fast pointers

   Slow moves one step, fast moves two steps; they will meet inside the loop, returning true.

3. Final Answer:

   Circular -> Option A

4. Quick Check:

   Loop detected by slow-fast pointers = Circular [OK]

Hint: Loop causes slow and fast pointers to meet [OK]

Common Mistakes:

• Ignoring the loop created by third->next
• Assuming no loop means Not Circular
• Confusing pointer assignments

4. Identify the error in the following code that attempts to detect if a linked list is circular:

bool isCircular(Node* head) {
  Node* slow = head;
  Node* fast = head;
  while(fast != nullptr) {
    slow = slow->next;
    fast = fast->next->next;
    if(slow == fast) return true;
  }
  return false;
}

medium

A. The loop condition should check fast->next != nullptr to avoid null pointer dereference

B. The slow pointer should move two steps instead of one

C. The function should return false inside the loop

D. The function should initialize slow and fast to nullptr

Solution

1. Step 1: Check loop condition safety

   fast->next->next is accessed, so fast->next must not be nullptr to avoid crash.

2. Step 2: Correct loop condition

   Change while(fast != nullptr) to while(fast != nullptr && fast->next != nullptr) to prevent null pointer dereference.

3. Final Answer:

   The loop condition should check fast->next != nullptr to avoid null pointer dereference -> Option A

4. Quick Check:

   Check fast and fast->next to avoid crash [OK]

Hint: Always check fast->next before fast->next->next [OK]

Common Mistakes:

• Not checking fast->next leads to runtime error
• Moving slow pointer incorrectly
• Returning false too early

5. You have a singly linked list where the last node might point back to any node, making it circular. How would you modify the list to remove the circular link and make it linear again?

hard

A. Set head to nullptr to clear the list

B. Delete all nodes after the head node

C. Reverse the linked list to break the loop

D. Detect the loop using slow and fast pointers, then find the node before the loop start and set its next to nullptr

Solution

1. Step 1: Detect loop using slow and fast pointers

   Use two pointers moving at different speeds; if they meet, a loop exists.

2. Step 2: Find the node before loop start and break the loop

   Reset one pointer to head, move both one step until they meet; the node before meeting point is loop end. Set its next to nullptr.

3. Final Answer:

   Detect the loop using slow and fast pointers, then find the node before the loop start and set its next to nullptr -> Option D

4. Quick Check:

   Find loop start, break link to remove circular [OK]

Hint: Find loop start, set previous node's next to nullptr [OK]

Common Mistakes:

• Deleting nodes unnecessarily
• Reversing list does not remove loop
• Setting head to nullptr loses entire list