在完成“计算字符串的相似度（VB2005）——思索之一”之后，照例对程序进行了一番测试。第一次找了两个相似的字符串，长度分别为15和17。速度和结果都比较满意。这也印证了算法的正确性。第二次找了两个相似的字符串，长度分别为1500和1507。嗯，直接跳出错误，说是堆栈错误。实际上是由于递归嵌套出了问题。采用递归算法，只是理论上有效，便于理解，实际应用中会出现各种限制。如本例，嵌套约1000层的时候就超过了系统的限制。必须想一个解决之道。
仔细观察，可以发现用数学性的语言描述就是
F（n，m）=G（F（n，m），F（n+1，m），F（n，m+1））
这个可以简化为递推，由于递推可以放在一个函数内，就解决了系统的递归限制。代码赋予其后。用的是VB2005
Public Class clsCalculateStringDistanceEx
Implements IDistance
Private mStringA As String
Private mStringB As String
Private mIsSame As Boolean
Private mDic As Dictionary(Of String,Integer)
Private Function CalculateStringDistance(ByVal StartLower As Integer) As Integer
Dim LA As Integer = mStringA.Length
Dim LB As Integer = mStringB.Length
Dim i As Integer,j As Integer
Dim T1 As Integer,T2 As Integer,T3 As Integer
AddToDic(LA + 1,LB + 1,0)
For i = LA To StartLower Step -1
AddToDic(i,LA - i + 1)
Next
For i = LB To StartLower Step -1
AddToDic(LA + 1,i,LB - i + 1)
Next
For i = LA To StartLower Step -1
For j = LB To StartLower Step -1
If mStringA.Chars(i - 1) = mStringB.Chars(j - 1) Then
AddToDic(i,j,GetFromDic(i + 1,j + 1))
Else
T1 = GetFromDic(i + 1,j)
T2 = GetFromDic(i,j + 1)
T3 = GetFromDic(i + 1,j + 1)
AddToDic(i,Min(T1,T2,T3) + 1)
End If
Next
Next
Return GetFromDic(StartLower,StartLower)
End Function
Private Sub AddToDic(ByVal S1 As Integer,ByVal S2 As Integer,ByVal Value As Integer)
mDic.Add(S1 & "," & S2,Value)
End Sub
Private Function GetFromDic(ByVal S1 As Integer,ByVal S2 As Integer) As Integer
Return mDic(S1 & "," & S2)
End Function
Private Function Min(ByVal ParamArray M() As Integer) As Integer
Dim i As Integer,J As Integer
J = M(0)
For i = 1 To M.GetUpperBound(0)
If M(i) < J Then J = M(i)
Next
Return J
End Function
Public Function CalculateStringDistance() As Integer _
Implements IDistance.CalculateStringDistance
If mStringA.Length = 0 Then Return mStringB.Length
If mStringB.Length = 0 Then Return mStringA.Length
mDic = New Dictionary(Of String,Integer)
mIsSame = True
Dim i As Integer,j As Integer
For i = 1 To Min(mStringA.Length,mStringB.Length)
If mStringA.Chars(i - 1) <> mStringB.Chars(i - 1) Then
mIsSame = False
j = i
Exit For
End If
Next
If mIsSame = False Then
Return CalculateStringDistance(j)
Else
Return Math.Abs(mStringA.Length - mStringB.Length)
End If
End Function
Public ReadOnly Property DicCount() As Integer _
Implements IDistance.DicCount
Get
Return mDic.Count
End Get
End Property
Public Sub SetString(ByVal S1 As String,ByVal S2 As String) _
Implements IDistance.SetString
mStringA = S1
mStringB = S2
End Sub
End Class
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