Check if a linked list is Circular Linked List
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):
if linkedList.head is NULL:
return 0
temp = linkedList.head
while temp is not NULL:
temp = temp.next
if temp is linkedList.head:
return 1
if temp is NULL:
return 0
Algorithm Explanation
 Check if the linked list is empty. If it's empty, return 0 (not circular).
 Initialize
temp
to the head of the linked list.  Traverse the linked list using the
temp
pointer. Iftemp
becomes equal to the head again, return 1 (circular).  If
temp
becomes NULL, return 0 (not circular).
Code Solution

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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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