你可以这样拆分
Sub split_array()
Dim array1(1 To 2) As String
Dim array2(1 To 2) As String
Dim array3(1 To 2) As String
array1(1) = "Test1"
array1(2) = "Test2"
array2(1) = array1(1)
array3(1) = array1(2)
End Sub
但我怀疑这不是最好的方法。我认为使用 3 个(可能是长整数)变量来表示数组中的位置会做得更好。 1代表第一个元素,1代表最后一个元素,1代表中间元素。
Dim lLowerSearchElement As Long
Dim lUpperSearchElement As Long
Dim lMiddleSearchElement As Long
Dim array1(1 to 999) as string
lLowerSearchElement = 1
lUpperSearchElement = 999
lMiddleSearchElement = (lUpperSearchElement + lLowerSearchElement) / 2
然后,您可以检查元素是否等于、大于或小于中间元素并相应地继续。
还请记住,在尝试使用二进制搜索之前,您需要对数据进行排序,如果您了解递归调用,这将很有用。
您还需要严格测试您的实现,因为一个小错误可能会导致搜索无法正常工作。
22/08/13 编辑
我用于二分搜索的实现如下:
Function bCheckSamplePoint(ByRef lSamplePointArray() As String, ByRef bfound As Boolean, _
ByVal lSamplePoint As String) As Boolean
'byref used for the array as could be slow to keep copying the array, bFound is used by calling procedure
Dim lLowerSearchElement As Long
Dim lUpperSearchElement As Long
Dim lMiddleSearchElement As Long
bfound = False 'False until found
'Set initial limits of the search
lLowerSearchElement = 0
lUpperSearchElement = UBound(lSamplePointArray())
Do While lLowerSearchElement <= lUpperSearchElement And bfound = False
lMiddleSearchElement = (lUpperSearchElement + lLowerSearchElement) / 2
If StrComp(lSamplePointArray(lMiddleSearchElement), lSamplePoint, vbTextCompare) = -1 Then
' 'Must be greater than middle element
lLowerSearchElement = lMiddleSearchElement + 1
ElseIf (lSamplePointArray(lMiddleSearchElement) = lSamplePoint) Then
bfound = True
Else
'must be lower than middle element
lUpperSearchElement = lMiddleSearchElement - 1
End If 'lSamplePointArray(lmiddlesearchlelemnt) < lSamplePoint
Loop 'While lLowerSearchElement <= lUpperSearchElement
ErrorExit:
bCheckSamplePoint = bReturn
Exit Function
正如您所见,这种二分搜索只是检查是否在字符串数组中找到了一个字符串,但它可以被修改用于其他目的。