How to write a String Palindrome code

Question

Hello,

How can I write a String Palindrome code in VB 2008  ? I found code in java which is given below but its using reverse function to reverse the string, Do we have such function in vb 2008 ?

http://codesgeek.com/

Please suggest most efficient method to check String palindrome in vb 2008.

Answer 1

StrReverse()

Dim word, rev As String 
word = "abcdefg"
rev = StrReverse(word)
If word = rev Then  'word is a palindrome
    

---

Solitaire

Answer 2

Hello,

How can I write a String Palindrome code in VB 2008  ? I found code in java which is given below but its using reverse function to reverse the string, Do we have such function in vb 2008 ?

http://codesgeek.com/

Please suggest most efficient method to check String palindrome in vb 2008.

I think the most efficient method would be to create one string with the text you want to check as a palindrome. Then another string with that text reversed. Then test if the first string.ToUpper (to make all characters uppercase for the test as a palindrome that begins with an upper case character when reversed will end with the uppercase character and the strings will not equal each other then even though all characters are the same. For example Kayak would not equal kayaK in a string test unless all characters were converted to upper or lower case for the test.) equals the second string.ToUpper as a boolean.

---

La vida loca

Answer 3

You can go through the first half of the string comparing the characters to the ones coming from the other end:

Module Module1

    Function IsPalindrome(s As String) As Boolean
        If s.Length <= 1 Then Return True

        For i = 0 To s.Length \ 2
            If String.Compare(s(i), s(s.Length - i - 1), StringComparison.CurrentCultureIgnoreCase) <> 0 Then
                Return False
            End If
        Next

        Return True

    End Function

    Sub Main()
        Console.WriteLine(IsPalindrome("Kayak")) ' outputs "True"
        Console.ReadLine()

    End Sub

End Module

It is a little bit more awkward if you want to ignore punctuation: "Madam, I'm Adam" is often considered to be a palindrome.

HTH,

Andrew

Answer 4

Very nice Andrew.  :)

Option Strict On

Module Module1
    Sub Main()
        Do
            Console.WriteLine("Enter a line of text (""exit"" to quit)")
            Console.Write(":>")
            Dim word As String = Console.ReadLine
            If word.ToLower = "exit" Then Exit Do
            If IsPalindrome(word) Then
                Console.WriteLine("That text is a Palindrome!")
            Else
                Console.WriteLine("That text is not a Palindrome.")
            End If
            Console.WriteLine()
        Loop
    End Sub

    Public Function IsPalindrome(value As String) As Boolean
        If value.Length < 2 Then Return True
        Dim chrs = From c As Char In value Let cl = Char.ToLower(c)
                   Where Char.GetUnicodeCategory(cl) = Globalization.UnicodeCategory.LowercaseLetter
        Dim count As Integer = chrs.Count
        For i As Integer = 0 To CInt(count / 2)
            If Not chrs(i).cl = chrs(count - i - 1).cl Then Return False
        Next
        Return True
    End Function
End Module

---

Reed Kimble - "When you do things right, people won't be sure you've done anything at all"

Answer 5

Very nice Andrew.  :)

Aww, why do you have to go and say that when I'm going to point out your code doesn't compile with Option Strict On? :(

And it throws an exception for value.Length < 2.

So, it's your turn to find something wrong with my idea to ignore punctuation :)

Function IsPalindrome(s As String) As Boolean
    If s.Length <= 1 Then Return True

    s = Text.RegularExpressions.Regex.Replace(s, "[\W]", "")

    If s.Length <= 1 Then Return True

    For i = 0 To s.Length \ 2
        If String.Compare(s(i), s(s.Length - i - 1), StringComparison.CurrentCultureIgnoreCase) <> 0 Then
            Return False
        End If
    Next

    Return True

End Function

--
Andrew

Answer 6

Making use of Andrew's Regex statement and the StrReverse function, here is my updated code.  It includes a separate teststring which is altered, preserving the original string.

        Dim mystring, teststring As String
        Console.Write("Enter your string:  ")
        mystring = Console.ReadLine()
        teststring = mystring           'preserve the original string
        teststring = Text.RegularExpressions.Regex.Replace(teststring, "[\W]", "")  'all punctuation & spaces removed
        teststring = teststring.ToLower()
        If mystring <> teststring Then
            Console.WriteLine("String was altered:  " & teststring)
        End If
        If teststring = StrReverse(teststring) Then
            Console.WriteLine("Your string is a palindrome.")
        Else
            Console.WriteLine("Your string is not a palindrome.")
        End If
        Console.ReadLine()

---

Solitaire

Answer 7

LOL those are easy fixes.  =P

---

Reed Kimble - "When you do things right, people won't be sure you've done anything at all"

Answer 8

...

So, it's your turn to find something wrong with my idea to ignore punctuation :)

Nope, can't find any.  ;)

Looks like Regex is faster than LINQ as well.  I wonder if manually writing the loop would beat LINQ...

---

Reed Kimble - "When you do things right, people won't be sure you've done anything at all"

Answer 9

...

So, it's your turn to find something wrong with my idea to ignore punctuation :)

Nope, can't find any.  ;)

Looks like Regex is faster than LINQ as well.  I wonder if manually writing the loop would beat LINQ...

---

Reed Kimble - "When you do things right, people won't be sure you've done anything at all"

LOL, oh yeah it does:

Public Function IsPalindrome3(value As String) As Boolean
    If value.Length < 2 Then Return True
    Dim chrs(value.Length - 1) As Char
    Dim index As Integer
    For Each c As Char In value
        If Char.IsLetter(c) Then
            chrs(index) = Char.ToLower(c)
            index += 1
        End If
    Next
    Dim count As Integer = index
    For i As Integer = 0 To CInt(count / 2)
        If Not chrs(i) = chrs(count - i - 1) Then Return False
    Next
    Return True
End Function

---

Reed Kimble - "When you do things right, people won't be sure you've done anything at all"

Answer 10

Whoops, I had the terms backward in that LINQ statement so it was evaluating too many items, that's why it was slower than regex.  The statement should have been:

Dim chrs = From c As Char In value
           Where Char.IsLetter(c)
           Let cl = Char.ToLower(c)

The LINQ solution pays a hefty price for its first execution, but all calls after that average a little faster than the Regex solution.

However, its all really moot since the manual solution is an order of magnitude faster than any of the others. =P

---

Reed Kimble - "When you do things right, people won't be sure you've done anything at all"

Answer 11

Very nice Andrew.  :)

Aww, why do you have to go and say that when I'm going to point out your code doesn't compile with Option Strict On? :(

And it throws an exception for value.Length < 2.

So, it's your turn to find something wrong with my idea to ignore punctuation :)

Function IsPalindrome(s As String) As Boolean
    If s.Length <= 1 Then Return True

    s = Text.RegularExpressions.Regex.Replace(s, "[\W]", "")

    If s.Length <= 1 Then Return True

    For i = 0 To s.Length \ 2
        If String.Compare(s(i), s(s.Length - i - 1), StringComparison.CurrentCultureIgnoreCase) <> 0 Then
            Return False
        End If
    Next

    Return True

End Function

--
Andrew

That's weird. How could it not compile with Option Strict On when Option Strict On is displayed in the code?

Obviously I'm missing something here although I've not brought the code up in VS.

---

La vida loca

Answer 12

Option Strict On

Imports System.Text

Public Class Form1

    Dim TimeIt As New Stopwatch

    Private Sub Form1_Load(sender As Object, e As EventArgs) Handles MyBase.Load
        Me.CenterToScreen()
    End Sub

    Private Sub Button1_Click(sender As Object, e As EventArgs) Handles Button1.Click
        TimeIt.Restart()
        If TextBox1.Text <> "" Then
            Dim Test As Boolean = IsThisAPalindrome(TextBox1.Text)
            If Test = True Then
                Label1.Text = "This " & ChrW(34) & TextBox1.Text & ChrW(34) & " is a palindrome."
            ElseIf Test = False Then
                Label1.Text = "This " & ChrW(34) & TextBox1.Text & ChrW(34) & " is not a palindrome."
            End If
        End If
        Label2.Text = "Answer found in " & TimeIt.ElapsedTicks.ToString & " millionths of a second."
    End Sub


    Private Function IsThisAPalindrome(ByVal text As String) As Boolean
        Dim Result As Boolean = False
        Dim stringBuilder = New StringBuilder()

        For i = 0 To text.Count - 1
            If Char.IsPunctuation(text(i)) = False Then
                stringBuilder.Append(text(i))
            End If
        Next

        If stringBuilder.ToString.ToUpper = StrReverse(stringBuilder.ToString.ToUpper) Then Result = True

        Return Result
    End Function

End Class

---

La vida loca

Answer 13

That's weird. How could it not compile with Option Strict On when Option Strict On is displayed in the code? Obviously I'm missing something here although I've not brought the code up in VS.

When I posted my comment aboout that, it had "For i As Integer = 0 To count / 2", where, of course, "count / 2" gives a Double. The simple correction is to use "count \ 2" or "CInt(Math.Floor(count / 2))" (Math.Floor is to avoid banker's rounding and thus checking the middle elements twice for some odd values of count).

--
Andrew