A set of digital,2,3,4,5,4,4,4,2,2,1,1,6 [1], for example, divided into,1,1 [1], [3], [2,2,2],,4,4,4 [4], [5], [6] should how cent

CodePudding user response:

A train of thought to you
As a first step, sorting, as to with what method you decide, what effervescent what binary way too much, not introduced,
The second step is to do a loop, the current one and one is the same as in the same group, a set of different,

CodePudding user response:

Form to add a ListBox, Sorted attribute set to True,

 Option Explicit 
Private Declare Function SendMessagebyString Lib _ 
"User32" Alias "SendMessageA" (ByVal hWND As Long, _ 
ByVal wMsg As Long, ByVal wParam As Long, _ 
ByVal lParam As Long As String) 

Private Const LB_FINDSTRINGEXACT=& amp; H1A2 

Private Sub Command1_Click () 
Dim strSource As String, strItem () As a String, As Long As Long, I n 

,2,3,4,5,4,4,4,2,2,1,1,6 strSource="[1]" 

StrSource=Replace (Replace (Replace (strSource, "/", "")," ", "")," ", "") 
StrItem=Split (strSource, ", ") 

List1. Clear 
For I=0 To UBound (strItem) 
N=SendMessagebyString (List1. HWND LB_FINDSTRINGEXACT, 1, strItem (I)) 

If n=1 Then 
List1. AddItem strItem (I) 
List1. ItemData (List1. NewIndex)=1 
The Else 
List1. ItemData (n)=List1. ItemData (n) + 1 
End the If 
Next I 

ReDim strItem (List1. ListCount - 1) 
For I=0 To List1. ListCount - 1 
StrItem (I)="[" 
For n=1 To List1. ItemData (I) - 1 
StrItem (I)=strItem (I) & amp; List1. List (I) & amp; ", "
The Next n 
StrItem (I)=strItem (I) & amp; List1. List (I) & amp; "] 
"The Debug. Print strItem (I) 
Next I 
The Debug. Print the Join (strItem, ", ") 
End Sub 

Results:
,1,1 [1]
,2,2 [2]
[3]
,4,4,4 [4]
[5]
[6]
,2,2,1,1 [1], [2], [3], [4,4,4,4], [5], [6]

CodePudding user response:

 Option Explicit 

'engineering reference: Microsoft Scripting Runtime 

Private Type MembList 
BuffSize As Long 
MembNum As Long 
DataBuff () As Long 
End Type 

Private Type GroupInfo 
ListSize As Long 
GroupNum As Long 
GroupList () As MembList 
End Type 

'* * * press "subscript starting from 0" processing array * * * 
Private stcDataGroup As GroupInfo 'grouping information management variable 

'* * * * * * * * * * * * * * * * * * * * * * * * * * * * * 
'* * will be a set of data grouping 
'* * * * * * * * * * * * * * * * * * * * * * * * * * * * * 
Private Sub Grouping (DataList () As Long) 
Dim objDict As a Dictionary 
Dim p As Long, n As Long, v As Long 
Dim As Long I 
'initialize grouping information 
StcDataGroup. ListSize=16 
StcDataGroup. GroupNum=0 
ReDim stcDataGroup. GroupList (stcDataGroup. ListSize - 1) 
'
 start processing dataThe Set objDict=New Dictionary 
For I=0 To UBound (DataList) 
V=DataList (I) 
If (objDict. The Exists (v)) Then 
P=objDict. Item (v) 
With stcDataGroup. GroupList (p) 
P=. MembNum 
If (p=. BuffSize) Then 
N=p + 4 'most larger, extend the length of the list, if the list length YiQu slightly larger value, 
ReDim Preserve. DataBuff (n - 1) 
. BuffSize=n 
End the If 
DataBuff (p)=v 
. MembNum=p + 1 '
End With 
The Else 
P=stcDataGroup. GroupNum 
ObjDict. Add v p 
'n=stcdatagroup. GroupList (p). MembNum 
If (p=stcDataGroup. ListSize) Then 'add groups for 
N=p + 8 
'expansion of eight at a time, can according to your need to change, group number, YiQu slightly larger value, 
'every time extension number is big, efficiency high, 
StcDataGroup. ListSize=n 
ReDim Preserve stcDataGroup. GroupList (n - 1) 
End the If 
'initialize new grouping 
With stcDataGroup. GroupList (p) 
N=16 'set the initial size in each group, packet length, is a larger value YiQu, 
. BuffSize=n 
ReDim. DataBuff (n - 1) 
. MembNum=1 
DataBuff (0)=v 
End With 
StcDataGroup. GroupNum=p + 1 
End the If 
Next 
ObjDict. RemoveAll 
End Sub 

'* * * * * * * * * * * * * * * * * * * * * * * * * * * * * 
'* * will be grouped data sample output (branch information obtaining sample) 
'* * * * * * * * * * * * * * * * * * * * * * * * * * * * * 
Private Sub ListGroup () 
Dim i& J& 

For I=0 To stcDataGroup. GroupNum - 1 
The Debug. Print "first" & amp; I + 1 & amp; "Set of data: 
"For j=0 To stcDataGroup GroupList (I). MembNum - 1 
The Debug. Print stcDataGroup. GroupList (I). DataBuff (j); 
Next 
The Debug. Print 
Next 
End Sub 

Private Sub Command1_Click () 
Dim aData () As Long 
Dim i& , sTxtBuf $() 

,2,3,4,5,4,4,4,2,2,1,1,6 sTxtBuf=Split (" 1 ", ", ") 
ReDim aData (UBound (sTxtBuf)) 
For I=0 To UBound (sTxtBuf) 
AData (I)=sTxtBuf (I) 
Next 
Call the Grouping (aData) 'data packet 
Call ListGroup 'output sample 
End Sub 

CodePudding user response:

Cattle are so many ha ha technology, Daniel

CodePudding user response:

Just suddenly thought of a problem, that is: "in each group, members are all the same",
MembList as a result, the data type is not necessary to use an array to write to all members, as long as the record what value is, how many went,
Such a scheme, but also can simplify data management, and improve operational efficiency,

So my code can simplify:

 Option Explicit 

'engineering reference: Microsoft Scripting Runtime 

Private Type MembList 
MembNum As Long 
The Value As Long 
End Type 

Private Type GroupInfo 
ListSize As Long 
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