计算字符串的相似度(VB2005)——思索之三

前端之家收集整理的这篇文章主要介绍了计算字符串的相似度(VB2005)——思索之三前端之家小编觉得挺不错的,现在分享给大家,也给大家做个参考。

在参阅本篇文章之前,建议先参阅以下两篇文章

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

在第二篇文章中的代码完成之后,照例还是对代码测试了一番。还是用两个相似的字符串,长度分别为15001507,结果能出来,但是效率差了点。在笔者的电脑上用了6秒中左右。仅仅是比较文本,就要6秒钟,比较难以接受,而且从代码看时间复杂度和空间复杂度都是On2)。

必须得改进!!!

在看了代码之后,发现代码运行速度慢可能出现在两个地方。一个是mDic对象,用的是Dictionary对象,在运行中反复读取和存储可能会影响速度,如果改为用数组可能效果会好点。哪位对这个有研究的同道,望不吝赐教。一个是String对象的CharsIndex)的方法。可能在每次执行到这一步时,会先把字符串转化为字符数组再返回一个字符,或者是遍历这个字符串,返回一个字符。对于本例中,大约需要执行1500×1500次,等于反复遍历,时间就浪费了。建议一开始就转化为字符数组,等到比较时就不需要遍历或转化了。

按照这两个思路对代码进行了修改,然后测试。效果很满意,本例测试几乎就是一瞬间。

现将代码赋予其后,用的是VB2005

Public Class clsCalculateStringDistanceEx2

Implements IDistance

Private mStringA() As Char

Private mStringB() As Char

Private mLenA As Integer

Private mLenB As Integer

Private mIsSame As Boolean

Private mDic(,) As Integer

Private Function CalculateStringDistance( _

ByVal StartLower As Integer,_

ByVal EndUper As Integer) As Integer

Dim i As Integer,j As Integer

Dim T1 As Integer,T2 As Integer,T3 As Integer

Dim jA As Integer = mLenA - StartLower - EndUper + 2

Dim jB As Integer = mLenB - StartLower - EndUper + 2

ReDim mDic(jA,jB)

mDic(jA,jB) = 0

For i = jA - 1 To 0 Step -1

mDic(i,jB) = jA - i

Next

For i = jB - 1 To 0 Step -1

mDic(jA,i) = jB - i

Next

For i = jA - 1 To 0 Step -1

For j = jB - 1 To 0 Step -1

If mStringA(i - 1 + StartLower) = _

mStringB(j - 1 + StartLower) Then

mDic(i,j) = mDic(i + 1,j + 1)

Else

T1 = mDic(i + 1,j)

T2 = mDic(i,j + 1)

T3 = mDic(i + 1,j + 1)

mDic(i,j) = Min(T1,T2,T3) + 1

End If

Next

Next

Return mDic(0,0)

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 mLenA = 0 Then Return mLenB

If mLenB = 0 Then Return mLenA

mIsSame = True

Dim i As Integer,j1 As Integer,j2 As Integer

For i = 1 To Min(mLenA,mLenB)

If mStringA(i - 1) <> mStringB(i - 1) Then

mIsSame = False

j1 = i

Exit For

End If

Next

If mIsSame = True Then Return Math.Abs(mLenA - mLenB)

For i = 1 To Min(mLenA,mLenB)

If mStringA(mLenA - i) <> mStringB(mLenB - i) Then

mIsSame = False

j2 = i

Exit For

End If

Next

If mIsSame = True Then Return Math.Abs(mLenA - mLenB)

Return CalculateStringDistance(j1,j2)

End Function

Public ReadOnly Property DicCount() As Integer _

Implements IDistance.DicCount

Get

Return mDic.Length

End Get

End Property

Public Sub SetString(ByVal S1 As String,ByVal S2 As String) _

Implements IDistance.SetString

mStringA = S1.tocharArray

mStringB = S2.tocharArray

mLenA = S1.Length

mLenB = S2.Length

End Sub

End Class

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