求算法——关于字符串映射替换的有关问题

求算法——关于字符串映射替换的问题
比如：原字符串“（A1+D3）*A11”

有如下映射关系 A1 -> Var1
D3 -> Var2
A11 -> Var3

转换后的字符串应为“(Var1+Var2)*Var3”

请问擅长于算法的朋友，应该实现上面的转换？谢谢！！

------解决方案--------------------
属于基本的算术式解析，给你一段四则运算解析代码示例：

VB code


Option Base 1
Dim tt(20) As String          '存放按+,-号分解出来的式子,只设定了20个

Private Sub Command1_Click()
    Text2.Text = handle(Text1.Text)
End Sub

Private Function handle(a)                  '按括号分解式子
    Dim th, th2 As String                      '存放按括号分解出来的式子
    Do While True
        n1 = InStr(a, "(")                          '查找
        n2 = InStr(a, ")")
        Select Case True
            Case n1 = 0 And n2 = 0                      '没有括号
                Exit Do
                Case n1 > 0 And n2 = 0                      '有 "(" 没 ")"
                MsgBox "表达式错误"
                Exit Function
            Case n2 > 0 And n1 = 0                      '有 ") 没 "("
                MsgBox "表达式错误"
                Exit Function
            Case n1 > n2                                '出现") ("
                MsgBox "表达式错误"
                Exit Function
            Case Else
                Do                 '如果出现这种情况"((( )))",则取最里面的括号开始分解
                    zl = n1
                    n1 = InStr(n1 + 1, a, "(")
                Loop Until n1 > n2 Or n1 = 0
                th = Mid(a, zl, n2 - zl + 1)                '存放分解出来的式子
                th2 = Mid(th, 2, Len(th) - 2)               '去括号
                a = Replace(a, th, handle1(th2)) '计算分解出来的式子,并把结果替换进去
        End Select
    Loop
    handle = handle1(a)                         '得到最后结果
End Function

Private Function handle1(b)                 '把handle3的每个结果相加
    n = handle2(b)
    s = 0                                       '累加数初值
    For i = 1 To n
        s = s + handle3(tt(i))                      '计算各乘除式的值并累加
    Next i
    handle1 = s                                 '得到结果
End Function
  
Private Function handle2(t)                 '按+,-分解式子
    '添以"+"开头,方便处理
    If Left(t, 1) <> "+" And Left(t, 1) <> "-" Then t = "+" & t
    k = 0                                      '用已累计分解出来的式子数
    Do While t <> ""
        n1 = InStr(2, t, "+")                       '查找
        n2 = InStr(2, t, "-")
        k = k + 1
        Select Case True
            Case n1 = 0 And n2 = 0                      '无+、-号
                tt(k) = t                                   '取剩余部分
                t = ""
            Case n2 = 0 Or (n1 < n2 And n1 > 0)         '无-号或先遇到+号
                tt(k) = Left(t, n1 - 1)                     '从该+号开始截取
                t = Mid(t, n1)
            Case n1 = 0 Or (n1 > n2 And n2 > 0)         '无+号或先遇到-号
                tt(k) = Left(t, n2 - 1)                     '从该-号开始截取
                t = Mid(t, n2)
        End Select
    Loop
    handle2 = k                                 '分解出来的式子个数
End Function

Private Function handle3(x)              '计算handle2分解出来的式子
    f = Left(x, 1)                              '取出符号+或-
    x = "*" & Mid(x, 2)                         '添以"*"开头,方便处理
    Sum = 1                                     '累乘/除的初值
    Do While x <> ""
        n1 = InStr(2, x, "*")                       '从第2个字符开始查找
        n2 = InStr(2, x, "/")
        Select Case True
            Case n1 = 0 And n2 = 0                      '无乘除号
                y = x
                x = ""
            Case n2 = 0 Or (n1 < n2 And n1 > 0)         '无/号或先遇到*号
                y = Left(x, n1 - 1)                         '取至*号前
                x = Mid(x, n1)
            Case n1 = 0 Or (n1 > n2 And n2 > 0)         '无*号或先遇到/号
                y = Left(x, n2 - 1)                         '取至/号前
                x = Mid(x, n2)
        End Select
        If Left(y, 1) = "*" Then                 '根据式子前面的符号进行运算
            Sum = Sum * Val(Mid(y, 2))
        Else
            Sum = Sum / Val(Mid(y, 2))
        End If
    Loop
    handle3 = Val(f & Str(Sum))                 '加上正负号
End Function

求算法——关于字符串映射替换的有关问题

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