Subtraction of alternate nodes of linked list in vb.net
Vb program for Subtraction of alternate nodes of linked list. Here problem description and other solutions.
' Include namespace system
Imports System
' Vb.net program for
' Subtraction of the alternate nodes of linked list
' Linked list node
Public Class LinkNode
Public data As Integer
Public [next] As LinkNode
Public Sub New(ByVal data As Integer)
Me.data = data
Me.next = Nothing
End Sub
End Class
public Class SingleLL
Public head As LinkNode
Public tail As LinkNode
Public Sub New()
' Set initial value
Me.head = Nothing
Me.tail = Nothing
End Sub
Public Sub insert(ByVal data As Integer)
Dim node As LinkNode = New LinkNode(data)
if (Me.head Is Nothing) Then
' Add first node
Me.head = node
Else
' Add node at the end position
Me.tail.[next] = node
End IF
' New last node
Me.tail = node
End Sub
' Display linked list element
Public Sub display()
if (Me.head Is Nothing) Then
Return
End If
Dim temp As LinkNode = Me.head
' iterating linked list elements
while (temp IsNot Nothing)
Console.Write(" " + temp.data.ToString() + " →")
' Visit to next node
temp = temp.[next]
End While
Console.WriteLine(" NULL")
End Sub
' Find the subtraction of all alternate
' nodes in linked list
Public Sub alternateSubtraction()
' Define resultant variables
Dim result As Integer = 0
Dim counter As Integer = 0
' Start to first node of linked list
Dim temp As LinkNode = Me.head
' iterating linked list elements
while (temp IsNot Nothing)
if (counter Mod 2 = 0) Then
' When get alternate node
if (result = 0) Then
result = temp.data
Else
result = result - temp.data
End IF
End If
' Node counter
counter += 1
' Visit to next node
temp = temp.[next]
End While
Console.WriteLine(" Alternate nodes subtraction : " +
result.ToString())
End Sub
Public Shared Sub Main(ByVal args As String())
Dim sll As SingleLL = New SingleLL()
' Add node in linked list
' 4 → 7 → 2 → 9 → 1 → 3 → 4 → 3 → 6 → NULL
sll.insert(4)
sll.insert(7)
sll.insert(2)
sll.insert(9)
sll.insert(1)
sll.insert(3)
sll.insert(4)
sll.insert(3)
sll.insert(6)
' Display of linked list nodes
sll.display()
' Test
sll.alternateSubtraction()
End Sub
End Class
Output
4 → 7 → 2 → 9 → 1 → 3 → 4 → 3 → 6 → NULL
Alternate nodes subtraction : -9
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