计算字符串的相似度（VB2005）——思索之四

参阅本文章之前，请先参阅以下文章

计算字符串的相似度（VB2005）——思索之一

计算字符串的相似度（VB2005）——思索之二

计算字符串的相似度（VB2005）——思索之三

在思索之三的程序完成之后，经测试，结果和速度都令人满意，稍显美中不足的是就是空间复杂度还是比较高，为O（S1×S2），当S1和S2都比较大的时候，可能会占用非常多的空间。

如何解决这个问题呢？

经过对计算过程的分析，我发现作为存储的二维矩阵，在每一个循环中，其实只有一行的数据参与了计算，之前的数据行就不再参与计算了。因此，从这个出发点入手，对代码进行了微调，将二维数组改为一维数组。经测试，结果和速度与之前思索之三中的代码没有差异。但空间复杂度少了很多，为O（S1）。

有意思的是，再对代码进行测试时，无意中发现了之前的代码中存在一个不起眼的逻辑错误。请参阅者自行考量。

现将代码赋予其后，用的是VB2005

Public Class clsDistance

Private mCharA() As Char

Private mCharB() As Char

Private mCharALen As Integer

Private mCharBLen As Integer

Public Sub New(ByVal StrA As String,ByVal StrB As String)

mCharA = StrA.ToCharArray

mCharB = StrB.ToCharArray

mCharALen = mCharA.Length

mCharBLen = mCharB.Length

End Sub

Public Function CacuDistance() As Integer

Dim i As Integer

If mCharALen = 0 Then Return mCharBLen

If mCharBLen = 0 Then Return mCharALen

Dim j As Integer = Min(mCharALen,mCharBLen) - 1

Dim tP1 As Integer,tP2 As Integer

tP1 = -1

tP2 = -1

For i = 0 To j

If mCharA(i) <> mCharB(i) Then

tP1 = i

Exit For

End If

Next

If tP1 = -1 Then Return Math.Abs(mCharALen - mCharBLen)

For i = 0 To j - tP1

If mCharA(mCharALen - i - 1) <> mCharB(mCharBLen - i - 1) Then

tP2 = i

If tP2 = -1 Then Return Math.Abs(mCharALen - mCharBLen)

Dim tA(mCharALen - tP1 - tP2) As Integer

For i = 0 To tA.GetUpperBound(0)

tA(i) = i

Dim tN1 As Integer,tN2 As Integer,tN3 As Integer

For i = 0 To mCharBLen - tP1 - tP2 - 1

tN1 = tA(0)

tN2 = tN1 + 1

For j = 1 To tA.GetUpperBound(0)

If mCharA(mCharALen - tP2 - j) = mCharB(mCharBLen - tP2 - i - 1) Then

tN3 = tN1

Else

tN3 = Min(tA(j),tN1,tN2) + 1

tA(j - 1) = tN2

tN2 = tN3

tN1 = tA(j)

tA(tA.GetUpperBound(0)) = tN2

Return tA(tA.GetUpperBound(0))

End Function

Public Function Min(ByVal ParamArray Num() As Integer) As Integer

Dim tN As Integer,i As Integer

If Num.Length = 0 Then Return Nothing

tN = Num(0)

For i = 1 To Num.GetUpperBound(0)

If Num(i) < tN Then tN = Num(i)

Return tN

End Function

End Class

（编辑：李大同）

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