1368 表达式转换

 描述 把中缀表达式转化为后缀表达式……
书上有详细讲解.
输入 第一行为T，表示有T组数据；
以下T行每行包含一个字符串，是中缀表达式。
输出 对于每组测试数据数据后缀表达式。
样例输入

1

A+(B-C/D)*E

样例输出

ABCD/-E*+

 
 
经典问题，用栈实现#include <iostream>
#include <string>
#include <stack>

using namespace std;

void Convert(const string & Infix, string & Postfix);

bool IsOperand(char ch);

bool TakesPrecedence(char OperatorA, char OperatorB);

int main(void)
{
char Reply;
int number;
scanf("%d",&number);
while(number--)

{
string Infix, Postfix; // local to this loop

cin >> Infix;

Convert(Infix, Postfix);

cout << Postfix << endl;

}

return 0;
}

/* Given: ch A character.
Task: To determine whether ch represents an operand (here understood
to be a single letter or digit).
Return: In the function name: true, if ch is an operand, false otherwise.
*/
bool IsOperand(char ch)
{
if (((ch >= 'a') && (ch <= 'z')) ||
((ch >= 'A') && (ch <= 'Z')) ||
((ch >= '0') && (ch <= '9')))
return true;
else
return false;
}

/* Given: OperatorA A character representing an operator or parenthesis.
OperatorB A character representing an operator or parenthesis.
Task: To determine whether OperatorA takes precedence over OperatorB.
Return: In the function name: true, if OperatorA takes precedence over
OperatorB.
*/
bool TakesPrecedence(char OperatorA, char OperatorB)
{
if (OperatorA == '(')
return false;
else if (OperatorB == '(')
return false;
else if (OperatorB == ')')
return true;
else if ((OperatorA == '^') && (OperatorB == '^'))
return false;
else if (OperatorA == '^')
return true;
else if (OperatorB == '^')
return false;
else if ((OperatorA == '*') || (OperatorA == '/'))
return true;
else if ((OperatorB == '*') || (OperatorB == '/'))
return false;
else
return true;

}

/* Given: Infix A string representing an infix expression (no spaces).
Task: To find the postfix equivalent of this expression.
Return: Postfix A string holding this postfix equivalent.
*/
void Convert(const string & Infix, string & Postfix)
{
stack<char> OperatorStack;
char TopSymbol, Symbol;
int k;

for (k = 0; k < Infix.size(); k++)
{
Symbol = Infix[k];
if (IsOperand(Symbol))
Postfix = Postfix + Symbol;
else
{
while ((! OperatorStack.empty()) &&
(TakesPrecedence(OperatorStack.top(), Symbol)))
{
TopSymbol = OperatorStack.top();
OperatorStack.pop();
Postfix = Postfix + TopSymbol;
}
if ((! OperatorStack.empty()) && (Symbol == ')'))
OperatorStack.pop(); // discard matching (
else
OperatorStack.push(Symbol);
}
}

while (! OperatorStack.empty())
{
TopSymbol = OperatorStack.top();
OperatorStack.pop();
Postfix = Postfix + TopSymbol;
}
}