在Permute函數中重複的問題

Question

我正在運行一個Permute函數，但是我遇到類似這樣的結果的問題，如何防止這種情況發生？

red|red|white 
white|red|red 

Public buffer As New List(Of String) 
Public Sub Permute(ByVal Root As String, ByVal Depth As Integer, ByVal Buffer1 As List(Of String)) 
    Dim data_array As String() = {"red", "blue", "white"} 
    For Each myStr As String In data_array 
     If Depth <= 1 Then 
      Buffer1.Add(Root & myStr) 
     Else 
      Permute(Root & myStr & ",", Depth - 1, Buffer1) 
     End If 
    Next 
End Sub 

Answer 1

Bjørn-Roger Kringsjå做此繁重在this great answer一些C++代碼轉換爲VB。

原始作品用於排列字符串字符。我修改它以使用字符串（我一直在做的事），並允許您指定分隔符。您可以返回一個List(Of String())或類似的號碼，並使用String.Join來自調用代碼。

Public NotInheritable Class Permutation 

    Public Shared Function Create(array As String(), sep As String) As List(Of String) 
     Return Permutation.Create(array, False, sep) 
    End Function 

    Public Shared Function Create(array As String(), sort As Boolean, 
            sep As String) As List(Of String) 
     If (array Is Nothing) Then 
      Throw New ArgumentNullException("array") 
     ElseIf ((array.Length < 0) OrElse (array.Length > 13)) Then 
      Throw New ArgumentOutOfRangeException("array") 
     End If 
     Dim list As New List(Of String) 
     Dim n As Integer = array.Length 
     Permutation.Permute(list, array, 0, array.Length, sep) 
     If (sort) Then 
      list.Sort() 
     End If 
     Return list 
    End Function 

    Private Shared Sub Permute(list As List(Of String), array As String(), 
           start As Int32, ndx As Int32, sep As String) 
     Permutation.Print(list, array, ndx, sep) 
     If (start < ndx) Then 
      Dim i, j As Integer 
      For i = (ndx - 2) To start Step -1 
       For j = (i + 1) To (ndx - 1) 
        Permutation.Swap(array, i, j) 
        Permutation.Permute(list, array, (i + 1), ndx, sep) 
       Next 
       Permutation.RotateLeft(array, i, ndx) 
      Next 
     End If 
    End Sub 

    Private Shared Sub Print(list As List(Of String), array As String(), 
          size As Int32, sep As String) 
     Dim tmp As New List(Of String) 
     If (array.Length <> 0) Then 
      For i As Integer = 0 To (size - 1) 
       tmp.Add(array(i)) 
      Next 
      list.Add(String.Join(sep, tmp)) 
     End If 
    End Sub 

    Private Shared Sub Swap(array As String(), i As Int32, j As Int32) 
     Dim tmp As String 
     tmp = array(i) 
     array(i) = array(j) 
     array(j) = tmp 
    End Sub 

    Private Shared Sub RotateLeft(array As String(), start As Int32, n As Int32) 
     Dim tmp As String = array(start) 
     For i As Integer = start To (n - 2) 
      array(i) = array(i + 1) 
     Next 
     array(n - 1) = tmp 
    End Sub 
End Class 

注意：我的mods將與Kringsjå先生的原始代碼很好地共存爲重載。用法：

Dim data = {"red", "blue", "white"} 

Dim combos = Permutation.Create(data, ", ") 

結果：

紅色，藍色，白色
紅，白，藍
藍色，紅色，白色
藍，白，紅
白色，紅色，藍色
白色，藍色，紅色

Answer 2

Wha你正在做的是選擇{ "red", "blue", "white" }Depth次之一。所以，如果你叫Permute("", 5, buffer)你會得到如下：

 
red,red,red,red,red 
red,red,red,red,blue 
... 
white,white,white,white,blue 
white,white,white,white,white 

這3到5個選項的力量。

如果你只想得到{ "red", "blue", "white" }所有可能的排列，然後做到這一點：

Public Function Permute(ByVal data As IEnumerable(Of String)) As IEnumerable(Of String) 
    If data.Skip(1).Any() Then 
     Return data.SelectMany(_ 
      Function (x) Permute(data.Except({ x })).Select(_ 
       Function (y) x & "," & y)) 
    Else 
     Return data 
    End If 
End Function 

...並調用它像這樣：

Dim results = Permute({ "red", "blue", "white" }) 

我得到的結果是：

 
red,blue,white 
red,white,blue 
blue,red,white 
blue,white,red 
white,red,blue 
white,blue,red 

這將是微不足道的，然後預先Root價值e對這些結果。