algorithm: find the middle of linked-list

Data-Structures/Linked-List/SinglyLinkedList.js

@@ -6,7 +6,7 @@
 * a singly linked list.
 */
 
-// Methods - size, head, addLast, addFirst, addAt, removeFirst, removeLast, remove, removeAt, indexOf, isEmpty, elementAt, get, clean
+// Methods - size, head, addLast, addFirst, addAt, removeFirst, removeLast, remove, removeAt, indexOf, isEmpty, elementAt, findMiddle, get, clean
 
 class Node {
  constructor (data) {
@@ -193,6 +193,19 @@ class LinkedList {
  return removedNode.data
  }
 
+ // Returns a reference to middle node of linked list
+ findMiddle () {
+ // If there are two middle nodes, return the second middle node.
+ let fast = this.headNode
+ let slow = this.headNode
+
+ while (fast != null && fast.next != null) {
+ fast = fast.next.next
+ slow = slow.next
+ }
+ return slow
+ }
+
  // make the linkedList Empty
  clean () {
  this.headNode = null

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

@@ -164,6 +164,29 @@ describe('SinglyLinkedList', () => {
  expect(list.size()).toBe(1)
  })
 
+ it('Middle node of linked list', () => {
+ const list = new LinkedList()
+ list.addFirst(1)
+
+ // Middle node for list having single node
+ expect(list.findMiddle().data).toEqual(1)
+
+ list.addLast(2)
+ list.addLast(3)
+ list.addLast(4)
+ list.addLast(5)
+ list.addLast(6)
+ list.addLast(7)
+
+ // Middle node for list having odd number of nodes
+ expect(list.findMiddle().data).toEqual(4)
+
+ list.addLast(10)
+
+ // Middle node for list having even number of nodes
+ expect(list.findMiddle().data).toEqual(5)
+ })
+
  it('Check Iterator', () => {
  const list = new LinkedList()