Question

我应该如何设置条件，只显示1,2、3、4或5个数字的变体。

Function SumUpRecursive(numbers As List(Of Integer), target As Integer, part As List(Of Integer), solutions As List(Of List(Of Integer))) As List(Of List(Of Integer))
    Dim s = part.Sum()
    If s = target Then
        ' MsgBox("sum(" & String.Join(",", part.[Select](Function(n) n.ToString()).ToArray()) & ")=" & target)
        solutions.Add(part)
    End If
    If s >= target Then
        Return Nothing
    End If
    For i As Integer = 0 To numbers.Count - 1
        Dim remaining = New List(Of Integer)()
        Dim n As Integer = numbers(i)
        For j As Integer = i + 1 To numbers.Count - 1
            remaining.Add(numbers(j))
        Next
        Dim part_rec = New List(Of Integer)(part)
        part_rec.Add(n)
        SumUpRecursive(remaining, target, part_rec, solutions)
    Next
    Return solutions
End Function

按钮：

 Dim numbers As Integer() = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15,
        16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34,
        35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53,
        54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72,
        73, 74, 75, 76, 77, 78, 79, 80}
    Dim target As Integer = 15
    Dim nums = SumUpRecursive((numbers.ToList()), target, New List(Of Integer), New List(Of List(Of Integer)))
    If nums Is Nothing Then
        MsgBox("Failure :(")
    Else
        TxtListScanTxt.Lines = nums.Select(Function(b) String.Join(", ", b)).ToArray()
    End If

输出：

1, 2, 3, 4, 5
1, 2, 3, 9
1, 2, 4, 8
1, 2, 5, 7
1, 2, 12
1, 3, 4, 7
1, 3, 5, 6
1, 3, 11
1, 4, 10
1, 5, 9
1, 6, 8
1, 14
2, 3, 4, 6
2, 3, 10
2, 4, 9
2, 5, 8
2, 6, 7
2, 13
3, 4, 8
3, 5, 7
3, 12
4, 5, 6
4, 11
5, 10
6, 9
7, 8
15

我只想显示3个组合（仅显示3位数字）。

3, 4, 8
3, 5, 7
4, 5, 6

以此类推...

我如何改进代码来做到这一点？仅计算3、4或5的组合，而不是全部。因为否则程序会崩溃，因此组合太多。大量。

Answer 1

添加一些重载来帮助

模块中的代码：

Public Function SumUpRecursive(numbers As IEnumerable(Of Integer), target As Integer, solutionLength As Integer) As IEnumerable(Of IEnumerable(Of Integer))
    Return SumUpRecursive(numbers, target).Where(Function(item) item.Count() <= solutionLength)
End Function
Public Function SumUpRecursive(numbers As IEnumerable(Of Integer), target As Integer) As IEnumerable(Of IEnumerable(Of Integer))
    Return SumUpRecursive(numbers, target, {}, New List(Of IEnumerable(Of Integer)))
End Function
Private Function SumUpRecursive(numbers As IEnumerable(Of Integer), target As Integer, part As IEnumerable(Of Integer), solutions As List(Of IEnumerable(Of Integer))) As IEnumerable(Of IEnumerable(Of Integer))
    Dim s = part.Sum()
    If s = target Then solutions.Add(part)
    If s >= target Then
        Return Nothing
    End If
    For i As Integer = 0 To numbers.Count - 1
        Dim remaining = New List(Of Integer)()
        Dim n As Integer = numbers(i)
        For j As Integer = i + 1 To numbers.Count - 1
            remaining.Add(numbers(j))
        Next
        Dim part_rec = New List(Of Integer)(part) From {n}
        SumUpRecursive(remaining, target, part_rec, solutions)
    Next
    Return solutions
End Function

代码形式：

Private Sub Button1_Click(sender As Object, e As EventArgs) Handles Button1.Click
    Dim low = 1, high = 80
    Dim numbers = Enumerable.Range(low, high - low + 1)
    Dim target = 15
    Dim nums = SumUpRecursive(numbers, target)
    If nums Is Nothing Then
        MsgBox("Failure :(")
    Else
        TxtListScanTxt.Lines = nums.Select(Function(b) String.Join(", ", b)).ToArray()
    End If
End Sub

还要注意，某些代码已得到简化，例如自动整数列表

Dim low = 1, high = 80
Dim numbers = Enumerable.Range(low, high - low + 1)

（如果您想排除一些数字，可以这样操作（很抱歉，我没有打扰您实际上是否从1到80的所有数字都在那里））：

Enumerable.Range(low, high - low + 1).Except({1, 2, 3, 5, 8, 13, 21, 34, 55})

因此，一旦您有重载，就没有理由调用大型递归函数，传递新列表了，这对于使用您的代码的其他人（您的教授？）来说确实令人困惑。您可以使用访问修饰符将其隐藏起来，然后简单地调用一个更简单的重载。

尽可能将类型更改为IEnumerable，而不是List。

但是将解决方案限制为最大数量的关键在于第一次过载，

SumUpRecursive(numbers, target).Where(Function(item) item.Count() <= solutionLength)

当然，如果您想限制为单个数量，可以这样做

SumUpRecursive(numbers, target).Where(Function(item) item.Count() = solutionLength)

，如果愿意，可以将其放在单独的重载中。最小/最大范围也可以。

我希望其中一些有用。

在VB.Net解决方案中将条件>等于<值

1 个答案: