algorithm: find the middle of linked-list

Data-Structures/Linked-List/SinglyLinkedList.js

@@ -193,6 +193,19 @@ class LinkedList {
     return removedNode.data
   }
 
+  // Returns a reference to middle node of linked list
+  MiddleOfLL () {
+    // 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 +164,32 @@ describe('SinglyLinkedList', () => {
     expect(list.size()).toBe(1)
   })
 
+  it('Middle node of linked list', () => {
+    const list = new LinkedList()
+    list.addFirst(1)
+
+    let MiddleNodeOfLinkedList = list.MiddleOfLL(list.headNode)
+    // Middle node for list having single node
+    expect(MiddleNodeOfLinkedList.data).toEqual(1)
+
+    list.addLast(2)
+    list.addLast(3)
+    list.addLast(4)
+    list.addLast(5)
+    list.addLast(6)
+    list.addLast(7)
+
+    MiddleNodeOfLinkedList = list.MiddleOfLL(list.headNode)
+    // Middle node for list having odd number of nodes
+    expect(MiddleNodeOfLinkedList.data).toEqual(4)
+
+    list.addLast(10)
+
+    MiddleNodeOfLinkedList = list.MiddleOfLL(list.headNode)
+    // Middle node for list having even number of nodes
+    expect(MiddleNodeOfLinkedList.data).toEqual(5)
+  })
+
   it('Check Iterator', () => {
     const list = new LinkedList()