Question

我不知道该怎么做。这是场景。我需要计算要删除的次数，以便字符串中没有任何连续的字符。例如

AAABBBCCC = 6个缺失。 2 on c，2 on b，2 on a使其不连续。

感谢您的帮助。我正在用VB.Net写作。

Answer 1

此方法适用于任何类型：

Shared Function CountConsecutiveDuplicates(Of T)(input As IEnumerable(Of T)) As Int32
    Dim count As Int32 = 0
    Dim comparer = EqualityComparer(Of T).Default

    ' optimization for lists and arrays: '
    Dim listT = TryCast(input, IList(Of T))
    If listT IsNot Nothing Then
        If listT.Count <= 1 Then Return 0
        For i As Int32 = 0 To listT.Count - 2
            If comparer.Equals(listT(i), listT(i + 1)) Then
                count += 1
            End If
        Next
        Return count
    End If

    If Not input.Any() Then Return 0
    Dim this As T = input.First()
    For Each item As T In input.Skip(1)
        If comparer.Equals(this, item) Then
            count += 1
        End If
        this = item
    Next
    Return count
End Function

用法：

Dim numDeletes = CountConsecutiveDuplicates("AAABBBCCC") '  6

你可以把它变成一个扩展方法：

Module GeneralExtensions
    <Extension()>
    Function CountConsecutiveDuplicates(Of T)(input As IEnumerable(Of T)) As Int32
        Dim count As Int32 = 0
        Dim comparer = EqualityComparer(Of T).Default

        ' optimization for lists and arrays: '
        Dim listT = TryCast(input, IList(Of T))
        If listT IsNot Nothing Then
            If listT.Count <= 1 Then Return 0
            For i As Int32 = 0 To listT.Count - 2
                If comparer.Equals(listT(i), listT(i + 1)) Then
                    count += 1
                End If
            Next
            Return count
        End If

        If Not input.Any() Then Return 0
        Dim this As T = input.First()
        For Each item As T In input.Skip(1)
            If comparer.Equals(this, item) Then
                count += 1
            End If
            this = item
        Next
        Return count
    End Function
End Module

然后你甚至可以这样使用它：

Dim numDeletes = "AAABBBCCC".CountConsecutiveDuplicates()

Answer 2

    Public Module Tester

    Public Sub Main()

        Dim _inStr As String = "AAABBBBCCCCC"
        Dim _reduced As Char() = _inStr.ToCharArray.Distinct().ToArray

        Console.WriteLine("{0} >> {1}", _inStr, New String(_reduced))

        For Each _c As Char In _reduced
            Console.WriteLine("Char: {0} remove: {1}", _c, _inStr.Where(Function(x) x = _c).Count - 1)
        Next

        Console.ReadKey()

    End Sub

End Module

输出：

AAABBBBCCCCC >> ABC
Char: A remove: 2
Char: B remove: 3
Char: C remove: 4

Answer 3

我不明白你想要如何计算删除量或者你打算如何处理结果，但简单地删除所有连续字符的简单解决方案可能如下所示：

Dim str as String = "AAABBBCCC"
For Each ch As Char In str
    str = str.Replace(ch, "")
    str = str & ch
Next

这将简单地返回＆＃34; ABC＆＃34;。如果需要，您可以随时添加一些变量和/或输出来计算每个字符的删除。

Answer 4

我认为你需要这样的东西，

    Dim myString As String = "AAAABBCCCDDDDD"
    Dim toDelete As List(Of Integer) = Nothing
    Dim intLoopIndex As Integer = 0
    Dim i As Integer

    For i = 0 To myString.Length - 1

        Do While myString(i).ToString = myString(i + 1).ToString
            toDelete.Add(i + 1)
        Loop
    Next

    Debug.WriteLine("You must delelte in total : " & toDelete.Count.ToString & " character(s)")

    For k = 0 To toDelete.Count - 1
        Debug.Write("Must delete the character from your string at position : " & toDelete(k).ToString)
    Next

试试吧！并玩：）

如何计算字符串中的连续字符

4 个答案: