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