Visual Basic 回文码

Question

我正在尝试创建一个应用程序，该应用程序将确定用户输入的字符串是否为回文。

是否可以不使用 StrReverse，可能使用 for next 循环。这就是我到目前为止所做的。

使用 StrReverse：

    Dim userInput As String = Me.txtbx1.Text.Trim.Replace(" ", "")
    Dim toBeComparedWith As String = StrReverse(userInput)

    Select Case String.Compare(userInput, toBeComparedWith, True)

        Case 0
            Me.lbl2.Text = "The following string is a palindrom"
        Case Else
            Me.lbl2.Text = "The following string is not a palindrom"

    End Select

不工作的一个：

    Dim input As String = TextBox1.Text.Trim.Replace(" ", "")
    Dim pallindromeChecker As String = input
    Dim output As String

    For counter As Integer = input To pallindromeChecker Step -1

        output = pallindromeChecker

    Next counter

    output = pallindromeChecker

    If output = input Then
        Me.Label1.Text = "output"
    Else
        Me.Label1.Text = "hi"
    End If

Answer 1

虽然使用字符串反转有效，但它不是最理想的，因为您要对字符串进行至少 2 次完整的迭代（因为字符串反转会创建一个字符串的副本，因为字符串在 .NET 中是不可变的）（加上额外的迭代为您的 @ 987654321@ 和 Replace 通话）。

但是考虑回文的基本属性：字符串的前半部分与字符串的后半部分相反。

检查回文的最佳算法只需要遍历输入字符串的一半——通过比较value[n] 和value[length-n] 来获得n = 0 to length/2。

在 VB.NET 中：

Public Shared Function IsPalindrome(value As String) As Boolean

    ' Input validation.
    If value Is Nothing Then Throw New ArgumentNullException("value")
    value = value.Replace(" ", "") // Note String.Replace(String,String) runs in O(n) time and if replacement is necessary then O(n) space.

    ' Shortcut case if the input string is empty.
    If value.Length = 0 Then Return False ' or True, depends on your preference

    ' Only need to iterate until half of the string length.
    ' Note that integer division results in a truncated value, e.g. (5 / 2 = 2)...
    '... so this ignores the middle character if the string is an odd-number of characters long.
    Dim max As Integer = value.Length - 1
    For i As Integer = 0 To value.Length / 2

        If value(i) <> value(max-i) Then
            ' Shortcut: we can abort on the first mismatched character we encounter, no need to check further.
            Return False
        End If

    Next i

    ' All "opposite" characters are equal, so return True.
    Return True

End Function