Skip to content
Merged
Show file tree
Hide file tree
Changes from 1 commit
Commits
File filter

Filter by extension

Conversations
Failed to load comments.
Loading
Jump to
Jump to file
Failed to load files.
Loading
Diff view
Diff view
Prev Previous commit
Next Next commit
Refactor: Added method in singly LL and its tests
  • Loading branch information
10kartik committed Sep 9, 2022
commit a70a3175a8b94e4b3a8539f291adb6d30330370f
27 changes: 0 additions & 27 deletions Data-Structures/Linked-List/MiddleOfLinkedList.js

This file was deleted.

13 changes: 13 additions & 0 deletions Data-Structures/Linked-List/SinglyLinkedList.js
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 () {
// 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
Expand Down
41 changes: 0 additions & 41 deletions Data-Structures/Linked-List/test/MiddleOfLinkedList.test.js

This file was deleted.

26 changes: 26 additions & 0 deletions Data-Structures/Linked-List/test/SinglyLinkedList.test.js
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)

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()

Expand Down