The problem involves checking whether a given singly linked list is a circular linked list or not. In a circular linked list, the last node's next pointer points back to the first node, creating a closed loop.

## Problem Statement

Given a singly linked list, we need to determine whether it's a circular linked list or not.

## Example

Consider the following linked list: 1 -> 2 -> 3 -> 4 -> 5. This is not a circular linked list because the last node's next pointer is NULL. However, if we connect the last node's next pointer to the first node, it becomes a circular linked list: 1 -> 2 -> 3 -> 4 -> 5 -> 1.

## Idea to Solve the Problem

To check if a linked list is a circular linked list, we can traverse the list while checking if the next pointer of the current node ever becomes NULL. If we encounter a NULL next pointer, it means the list is not circular. If we encounter a node whose next pointer points back to a previously visited node, it indicates that the list is circular.

## Pseudocode

Here's the pseudocode to check if a linked list is circular:

``````function isCircular(linkedList):
return 0

while temp is not NULL:
temp = temp.next
return 1
if temp is NULL:
return 0``````

## Algorithm Explanation

1. Check if the linked list is empty. If it's empty, return 0 (not circular).
2. Initialize `temp` to the head of the linked list.
3. Traverse the linked list using the `temp` pointer. If `temp` becomes equal to the head again, return 1 (circular).
4. If `temp` becomes NULL, return 0 (not circular).

## Time Complexity

The time complexity of checking whether a linked list is circular or not is O(n), where n is the number of nodes in the linked list. This is because we need to traverse the entire list to determine its circular nature.

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