如何使從數組元素的所有可能的總和組合在VB

Question

如果存在與元素的數組：1,2,3,4，程序應該返回另一個陣列的所有組合的總和：

1 
2 
3 
4 
3 (1+2) 
4 (1+3) 
5 (1+4) 
5 (2+3) 
6 (2+4) 
7 (3+4) 
6 (1+2+3) 
7 (1+2+4) 
8 (1+3+4) 
9 (2+3+4) 
10 (1+2+3+4)

Answer 1

我的想法是：

（pseudcode，我不;知道VB）

for(int i = 0; i < 4321; i++) 
{ 
    i mod 10 + // first from right digit 
    (int)((i mod 100)mod 10) // second, (?) 
    // etc 
    // sum up all 4 digit 
    // add to array 
} 

Answer 2

編碼您在您的評論中提到的算法，在僞VB代碼：

ReDim result(2^(Length of Array) - 1) 
for index = 0 to 2^(Length of Array) - 1 
    sum = 0 
    for counter = 0 to (Length of Array) - 1 
    If ((2^counter) And index) <> 0 Then 
     sum += Array(counter+1) 

    result(index) = sum 

Answer 3

這是我前段時間寫的一個函數，用於生成給定數組的所有可能的子集。它是通用的，所以它支持整數，雙打，字符串等

原來的C＃

public static List<T[]> CreateSubsets<T>(T[] originalArray) 
{ 
    List<T[]> subsets = new List<T[]>(); 

    for (int i = 0; i < originalArray.Length; i++) 
    { 
     int subsetCount = subsets.Count; 
     subsets.Add(new T[] { originalArray[i] }); 

     for (int j = 0; j < subsetCount; j++) 
     { 
      T[] newSubset = new T[subsets[j].Length + 1]; 
      subsets[j].CopyTo(newSubset, 0); 
      newSubset[newSubset.Length - 1] = originalArray[i]; 
      subsets.Add(newSubset); 
     } 
    } 

    return subsets; 
} 

而我只是轉換爲VB的版本。

Function CreateSubsets(Of T)(ByVal originalArray() As T) As List(Of T()) 

    Dim subsets As New List(Of T()) 

    For i As Integer = 0 To originalArray.Length - 1 

     Dim subsetCount As Integer = subsets.Count 
     subsets.Add(New T() {originalArray(i)}) 

     For j As Integer = 0 To subsetCount - 1 
      Dim newSubset(subsets(j).Length) As T 
      subsets(j).CopyTo(newSubset, 0) 
      newSubset(newSubset.Length - 1) = originalArray(i) 
      subsets.Add(newSubset) 
     Next 

    Next 

    Return subsets 

End Function 

它可以以這種方式被消費

Dim array() As Integer = {1, 2, 3, 4, 5} 
    Dim subsets As List(Of Integer()) = CreateSubsets(array) 

    For Each subset As Integer() In subsets 

     Dim sum As Integer = subset.Sum() 

    Next