refactor: Cycledetection.js and added tests

Data-Structures/Linked-List/CycleDetection.js

@@ -1,32 +1,24 @@
 /**
- * A LinkedList based solution for Detect a Cycle in a list
+ * A LinkedList based solution for Detecting a Cycle in a list.
  * https://en.wikipedia.org/wiki/Cycle_detection
  */
 
-function main () {
+function detectCycle (head) {
   /*
   Problem Statement:
   Given head, the head of a linked list, determine if the linked list has a cycle in it.
-
-  Note:
-  * While Solving the problem in given link below, don't use main() function.
-  * Just use only the code inside main() function.
-  * The purpose of using main() function here is to avoid global variables.
-
   Link for the Problem: https://leetcode.com/problems/linked-list-cycle/
   */
-  const head = '' // Reference to head is given in the problem. So please ignore this line
-  let fast = head
-  let slow = head
+  if (!head) { return false }
 
-  while (fast != null && fast.next != null && slow != null) {
+  let slow = head
+  let fast = head.next
+  while (fast && fast.next) {
+    if (fast === slow) { return true }
     fast = fast.next.next
     slow = slow.next
-    if (fast === slow) {
-      return true
-    }
   }
   return false
 }
 
-main()
+export { detectCycle }

Data-Structures/Linked-List/test/CycleDetection.test.js

@@ -0,0 +1,31 @@
+import { detectCycle } from '../CycleDetection'
+import { Node } from '../SinglyLinkedList'
+
+describe('Detect Cycle', () => {
+  it('should detect loop and return true', () => {
+    // Creating list and making a loop
+    const headNode = new Node(10)
+    headNode.next = new Node(20)
+    headNode.next.next = new Node(30)
+    headNode.next.next.next = new Node(40)
+    headNode.next.next.next.next = headNode
+    expect(detectCycle(headNode)).toEqual(true)
+  })
+
+  it('should not detect a loop and return false', () => {
+    // Case 0: When head is null, there is no loop.
+    expect(detectCycle(null)).toEqual(false)
+    const headNode = new Node(10)
+
+    // Case 1: List with single node doesn't have any loop
+    expect(detectCycle(headNode)).toEqual(false)
+
+    headNode.next = new Node(20)
+    headNode.next.next = new Node(30)
+    headNode.next.next.next = new Node(40)
+    headNode.next.next.next.next = new Node(50)
+
+    // Case 2: List not having any loops
+    expect(detectCycle(headNode)).toEqual(false)
+  })
+})