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如何计算字符串中零的数量，直到遇到非零 vb.net

发布于 2024-12-05 07:59:17 字数 79 浏览 3 评论 0原文

示例

如果用户输入 00001，计数将为 4

如果用户输入 0811，计数将为 1

我该怎么做？

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心如狂蝶 2024-12-12 07:59:17

最有效的方法是简单地循环遍历字符串的字符并计算每个 0 字符，直到读取到非 0 字符。

我不太使用 VB.NET，所以这只是一些粗略的 VBish 伪代码

Dim myString As String = "00001"
Dim count As Integer = 0
For Each c As Char In myString
  If C = "0"c Then
    count += 1
  Else
    Exit For
  End If
Next

The most efficient way is to simply loop through the characters of the string and count each character that is a 0 until your read a non 0 character.

I do not use VB.NET much, so this is just some rough VBish pseudo code

Dim myString As String = "00001"
Dim count As Integer = 0
For Each c As Char In myString
  If C = "0"c Then
    count += 1
  Else
    Exit For
  End If
Next

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请你别敷衍 2024-12-12 07:59:17

这是没有可见 for 循环的一行答案。

Module Module1

    Sub Main()
        Console.WriteLine(GetLeadingZeros("00001"))
        Console.WriteLine(GetLeadingZeros("0889"))
        Console.WriteLine(GetLeadingZeros("1"))
        Console.WriteLine(GetLeadingZeros("00101"))
        Console.WriteLine(GetLeadingZeros("11111"))
        Console.WriteLine(GetLeadingZeros("10001"))
        Console.ReadLine()
    End Sub

    Public Function GetLeadingZeros(ByVal input As String) As String
        Return input.Substring(0, input.IndexOf(input.SkipWhile(Function(e) e = "0")(0)))
    End Function
End Module

And here is the one line answer without a visible for loop.

Module Module1

    Sub Main()
        Console.WriteLine(GetLeadingZeros("00001"))
        Console.WriteLine(GetLeadingZeros("0889"))
        Console.WriteLine(GetLeadingZeros("1"))
        Console.WriteLine(GetLeadingZeros("00101"))
        Console.WriteLine(GetLeadingZeros("11111"))
        Console.WriteLine(GetLeadingZeros("10001"))
        Console.ReadLine()
    End Sub

    Public Function GetLeadingZeros(ByVal input As String) As String
        Return input.Substring(0, input.IndexOf(input.SkipWhile(Function(e) e = "0")(0)))
    End Function
End Module

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~没有更多了~