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分享问数组中含有重复字符串算法。
在C社区里问过的但C的指针实在太好了，VB里怎么办？
设：
Array("ab" ,"bc","ac","ab","sr","ac","ad") 大量的无规律字符串数组，求一函数若有两个字符串相同，则返回1，反之为0，如上例，因"ac"重复出现，所以返回1。
ps:各项的长度是相等的。
谢谢大家，我很期待大家会有什么样精彩的算法。
不限语言，不限方法，只要快就行（时间复杂度，空间复杂度……该死的名词见鬼去吧！）
设：
Array("ab" ,"bc","ac","ab","sr","ac","ad") 大量的无规律字符串数组，求一函数若有两个字符串相同，则返回1，反之为0，如上例，因"ac"重复出现，所以返回1。
ps:各项的长度是相等的。
谢谢大家，我很期待大家会有什么样精彩的算法。
不限语言，不限方法，只要快就行（时间复杂度，空间复杂度……该死的名词见鬼去吧！）
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excuseser 2004-07-21
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肯定不保证！欢迎 KiteGirl(小仙妹) 把方法贴出来啊。
KiteGirl 2004-07-20
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你确认肯定是两个半角英文小写字符串吗？这很重要的！如果你确定这一点，我有一个特别快的算法。
mmcgzs 2004-07-20
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佩服楼上的几位，牛。
rainstormmaster 2004-07-20
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用集合加上错误处理比较好
leolan 2004-07-20
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private function IsRepeat(v as variant) as boolean
on error resume next
dim o as new collection
dim i as integer
for i=lbound(v) to ubound(v)
o.add v(i),v(i)
if err then
IsRepeat = true
exit function
end if
next
end function
on error resume next
dim o as new collection
dim i as integer
for i=lbound(v) to ubound(v)
o.add v(i),v(i)
if err then
IsRepeat = true
exit function
end if
next
end function
northwolves 2004-07-20
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也可用集合处理：
Function existsame(ByRef x As Variant) As Integer
existsame = 0
On Error Resume Next
Dim i As Long, temp As New Collection
For i = 0 To UBound(x)
temp.Add x(i), x(i)
If temp.Count <= i Then
existsame = 1
Exit Function
End If
Next
End Function
Function existsame(ByRef x As Variant) As Integer
existsame = 0
On Error Resume Next
Dim i As Long, temp As New Collection
For i = 0 To UBound(x)
temp.Add x(i), x(i)
If temp.Count <= i Then
existsame = 1
Exit Function
End If
Next
End Function
northwolves 2004-07-20
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Option Explicit
Private Sub Command1_Click()
Dim x
x = Array("ab", "bc", "ac", "ab", "sr", "ac", "ad")
MsgBox existsame(x)
End Sub
Function existsame(ByRef x As Variant) As Integer
existsame = 0
Dim i As Long
For i = 0 To UBound(x)
If UBound(Filter(x, x(i), , vbBinaryCompare)) > 0 Then
existsame = 1
Exit Function
End If
Next
End Function
Private Sub Command1_Click()
Dim x
x = Array("ab", "bc", "ac", "ab", "sr", "ac", "ad")
MsgBox existsame(x)
End Sub
Function existsame(ByRef x As Variant) As Integer
existsame = 0
Dim i As Long
For i = 0 To UBound(x)
If UBound(Filter(x, x(i), , vbBinaryCompare)) > 0 Then
existsame = 1
Exit Function
End If
Next
End Function
northwolves 2004-07-20
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Function existsame(ByRef x As Variant) As Boolean
Dim i As Long
For i = 0 To UBound(x)
If UBound(Filter(x, x(i), , vbBinaryCompare)) > 0 Then
existsame = True
Exit Function
End If
Next
End Function
Dim i As Long
For i = 0 To UBound(x)
If UBound(Filter(x, x(i), , vbBinaryCompare)) > 0 Then
existsame = True
Exit Function
End If
Next
End Function
