algorithm: find the middle of linked-list

Data-Structures/Linked-List/MiddleOfLinkedList.js

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+/**
+ * A LinkedList based solution for finding middle node of linked list.
+ * https://afteracademy.com/blog/middle-of-the-linked-list
+ */
+class MiddleOfLL {
+  solution (head) {
+  /*
+    Problem Statement:
+    Given the head of a singly linked list, return the middle node of the linked list.
+    If there are two middle nodes, return the second middle node.
+
+    Link for the Problem: https://leetcode.com/problems/middle-of-the-linked-list/
+    */
+    let fast = head
+    let slow = head
+
+    if (head.next == null) { return head }
+
+    while (fast != null && fast.next != null) {
+      fast = fast.next.next
+      slow = slow.next
+    }
+    return slow
+  }
+}
+
+export { MiddleOfLL }

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

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+import { MiddleOfLL } from '../MiddleOfLinkedList'
+import { LinkedList } from '../SinglyLinkedList'
+
+describe('MiddleOfLinkedList', () => {
+  it('middle node Of linked list - even length ', () => {
+    const list = new LinkedList()
+    list.addFirst(1)
+    list.addLast(2)
+    list.addLast(3)
+    list.addLast(4)
+    list.addLast(5)
+    list.addLast(6)
+    list.addLast(7)
+
+    const MiddleNodeOfLinkedList = new MiddleOfLL().solution(list.headNode)
+    expect(MiddleNodeOfLinkedList.data).toEqual(4)
+  })
+
+  it('middle node of linked list - odd length ', () => {
+    const list = new LinkedList()
+    list.addFirst(10)
+    list.addLast(20)
+    list.addLast(30)
+    list.addLast(40)
+    list.addLast(50)
+    list.addLast(60)
+    list.addLast(70)
+    list.addLast(80)
+
+    const MiddleNodeOfLinkedList = new MiddleOfLL().solution(list.headNode)
+    expect(MiddleNodeOfLinkedList.data).toEqual(50)
+  })
+
+  it('middle node of linked list - length 1 ', () => {
+    const list = new LinkedList()
+    list.addFirst(100)
+
+    const MiddleNodeOfLinkedList = new MiddleOfLL().solution(list.headNode)
+    expect(MiddleNodeOfLinkedList.data).toEqual(100)
+  })
+})