algorithm: find the middle of linked-list #1096

Original file line number	Diff line number	Diff line change
Expand Up		@@ -193,6 +193,19 @@ class LinkedList {
		return removedNode.data
		}

	// Returns a reference to middle node of linked list
	MiddleOfLL () {
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	// 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) {
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	fast = fast.next.next
	slow = slow.next
	}
	return slow
	}

		// make the linkedList Empty
		clean () {
		this.headNode = null
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Original file line number	Diff line number	Diff line change
Expand Up		@@ -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)
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	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()

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