feat: FindMiddleNode in a Linked List Algorithm

Data-Structures/Linked-List/FindMiddleNode.js

@@ -0,0 +1,40 @@
+/*
+Problem Statement:
+Given a non-empty singly linked list, find the middle node.
+If there are two middle nodes, return the second middle node.
+
+Link for the Problem: https://leetcode.com/problems/middle-of-the-linked-list/
+*/
+
+class FindMiddleNode {
+  constructor() {
+    this.head = null
+  }
+
+  // Method to find and return the middle node of the linked list
+  findMiddle(head) {
+    let slow = head
+    let fast = head
+
+    while (fast !== null && fast.next !== null) {
+      slow = slow.next // Move slow by 1 step
+      fast = fast.next.next // Move fast by 2 steps
+    }
+
+    return slow // Slow will be at the middle node
+  }
+
+  // Convert list to array for easy checking in test cases
+  solutionToArray(node) {
+    const list = []
+    let currentNode = node
+    while (currentNode) {
+      list.push(currentNode.data)
+      currentNode = currentNode.next
+    }
+
+    return list
+  }
+}
+
+export { FindMiddleNode }

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

@@ -0,0 +1,35 @@
+import { FindMiddleNode } from '../FindMiddleNode.js'
+import { LinkedList } from '../SinglyLinkedList.js'
+
+describe('FindMiddleNode', () => {
+  it('Check Middle Node of Linked List', () => {
+    const list = new LinkedList()
+    list.addFirst(1)
+    list.addLast(2)
+    list.addLast(3)
+    list.addLast(4)
+    list.addLast(5)
+
+    const findMiddleNode = new FindMiddleNode()
+    const middleNode = findMiddleNode.findMiddle(list.headNode)
+
+    // Convert the node and its following nodes to an array for verification
+    expect(findMiddleNode.solutionToArray(middleNode)).toEqual([3, 4, 5])
+  })
+
+  it('Check Middle Node for Even Number of Elements', () => {
+    const list = new LinkedList()
+    list.addFirst(1)
+    list.addLast(2)
+    list.addLast(3)
+    list.addLast(4)
+    list.addLast(5)
+    list.addLast(6)
+
+    const findMiddleNode = new FindMiddleNode()
+    const middleNode = findMiddleNode.findMiddle(list.headNode)
+
+    // For even-length lists, the second middle node is returned
+    expect(findMiddleNode.solutionToArray(middleNode)).toEqual([4, 5, 6])
+  })
+})