堆栈应用4

中缀表达式到后缀表达式的转换要把表达式从中缀表达式的形式转换成用后缀表示法
表示的等价表达式

C# Code:

//using System;
class Class1
{
public static void Main()
{
System.Console.WriteLine("Hello World!");
//中缀 => 后缀表达式
string s = "( 1.9 + (20 + 41) / (25 * 11) - 3 ) * 2"; //中缀; //中缀
string S = ""; //后缀
char[] Operators = new char[s.Length];
int Top = -1;
for (int i = 0; i < s.Length; i++)
{
char C = s[i];
switch (C)
{
case ' ' : //忽略空格
break;
case '+' : //操作符
case '-' :
while (Top >= 0) //栈不为空时
{
char c = Operators[Top--]; //pop Operator
if (c == '(')
{
Operators[++Top] = c; //push Operator
break;
}
else
{
S = S + c;
}
}
Operators[++Top] = C; //push Operator
S += " ";
break;
case '*' : //忽略空格
case '/' :
while (Top >= 0) //栈不为空时
{
char c = Operators[Top--]; //pop Operator
if (c == '(')
{
Operators[++Top] = c; //push Operator
break;
}
else
{
if (c == '+' || c == '-')
{
Operators[++Top] = c; //push Operator
break;
}
else
{
S = S + c;
}
}
}
Operators[++Top] = C; //push Operator
S += " ";
break;
case '(' :
Operators[++Top] = C;
S += " ";
break;
case ')' :
while (Top >= 0) //栈不为空时
{
char c = Operators[Top--]; //pop Operator
if (c == '(')
{
break;
}
else
{
S = S + c;
}
}
S += " ";
break;
default :
S = S + C;
break;

}
}
while (Top >= 0)
{
S = S + Operators[Top--]; //pop Operator
}

System.Console.WriteLine(S); //后缀

//后缀表达式计算
double[] Operands = new double[S.Length];
double x, y, v;
Top = - 1;
string Operand = "";
for (int i = 0; i < S.Length; i++)
{
char c = S[i];
if ((c >= '0' && c <= '9') || c == '.')
{
Operand += c;
}

if ((c == ' ' && Operand != "") || i == S.Length - 1)
{
Operands[++Top] = System.Convert.ToDouble(Operand) ; //push Operands
Operand = "";
}

if (c == '+' || c == '-' || c == '*' || c == '/')
{
if ((Operand != ""))
{
Operands[++Top] = System.Convert.ToDouble(Operand) ; //push Operands
Operand = "";
}
y = Operands[Top--]; //pop 双目运算符的第二操作数 (后进先出)注意操作数顺序对除法的影响
x = Operands[Top--]; //pop 双目运算符的第一操作数
switch (c)
{
case '+' :
v = x + y;
break;
case '-' :
v = x - y;
break;
case '*' :
v = x * y;
break;
case '/' :
v = x / y; // 第一操作数 / 第二操作数 注意操作数顺序对除法的影响
break;
default :
v = 0;
break;
}
Operands[++Top] = v; //push 中间结果再次入栈
}
}
v = Operands[Top--]; //pop 最终结果
System.Console.WriteLine(v);
System.Console.ReadLine();
}
}


Java Code:

class Class1
{
public static void main(String[] args)
{
System.out.println("Hello World!");
//中缀 => 后缀表达式
String s = "( 1.9 + (20 + 41) / (25 * 11) - 3 ) * 2"; //中缀
String S = ""; //后缀
char[] Operators = new char[s.length()];
int Top = -1;
for (int i = 0; i < s.length(); i++)
{
char C = s.charAt(i);
switch(C)
{
case ' ' :
break;
case '+' : //操作符
case '-' :
while (Top >= 0) //栈不为空时
{
char c = Operators[Top--]; //pop Operator
if (c == '(')
{
Operators[++Top] = c; //push Operator
break;
}
else
{
S = S + c;
}
}
Operators[++Top] = C; //push Operator
S += " ";
break;
case '*' : //操作符
case '/' :
while (Top >= 0) //栈不为空时
{
char c = Operators[Top--]; //pop Operator
if (c == '(')
{
Operators[++Top] = c; //push Operator
break;
}
else
{
if (c == '+' || c == '-')
{
Operators[++Top] = c; //push Operator
break;
}
else
{
S = S + c;
}
}
}
Operators[++Top] = C; //push Operator
S += " ";
break;
case '(' : //操作符
Operators[++Top] = C;
S += " ";
break;
case ')' : //操作符
while (Top >= 0) //栈不为空时
{
char c = Operators[Top--]; //pop Operator
if (c == '(')
{
break;
}
else
{
S = S + c;
}
}
S += " ";
break;
default : //操作数
S = S + C;
break;
}
}
while (Top >= 0)
{
S = S + Operators[Top--]; //pop Operator
}

System.out.println(S); //后缀

//后缀表达式计算
double[] Operands = new double[S.length()];
double x, y, v;
Top = - 1;
String Operand = "";
for (int i = 0; i < S.length(); i++)
{
char c = S.charAt(i);
if ((c >= '0' && c <= '9') || c == '.')
{
Operand += c;
}

if ((c == ' ' && Operand != "") || i == S.length() - 1)
{
Operands[++Top] = java.lang.Double.parseDouble(Operand) ; //push Operands
Operand = "";
}

if (c == '+' || c == '-' || c == '*' || c == '/')
{
if ((Operand != ""))
{
Operands[++Top] = java.lang.Double.parseDouble(Operand) ; //push Operands
Operand = "";
}
y = Operands[Top--]; //pop 双目运算符的第二操作数 (后进先出)注意操作数顺序对除法的影响
x = Operands[Top--]; //pop 双目运算符的第一操作数
switch (c)
{
case '+' :
v = x + y;
break;
case '-' :
v = x - y;
break;
case '*' :
v = x * y;
break;
case '/' :
v = x / y; // 第一操作数 / 第二操作数 注意操作数顺序对除法的影响
break;
default :
v = 0;
break;
}
Operands[++Top] = v; //push 中间结果再次入栈
}
}
v = Operands[Top--]; //pop 最终结果 System.out.println(v); }}