将中缀表达式转换为后缀表达式

中缀表达式对人类来说是可读且可解的。我们可以轻松区分运算符的顺序，并且在计算数学表达式时可以使用括号来首先解决该部分。计算机不能轻易区分运算符和括号，因此需要后缀转换。

要将中缀表达式转换为后缀表达式，我们将使用栈数据结构。通过从左到右扫描中缀表达式，当我们获取到任何操作数时，只需将其添加到后缀形式中，对于运算符和括号，将它们按照优先级添加到栈中。

注意： 这里我们只考虑{+, −, ∗, /, ^}运算符，其他运算符被忽略。

输入和输出

Input:
The infix expression. x^y/(5*z)+2
Output:
Postfix Form Is: xy^5z*/2+

算法

infixToPostfix(infix)

输入− 中缀表达式。

输出− 转换为后缀表达式。

Begin
   initially push some special character say # into the stack
   for each character ch from infix expression, do
      if ch is alphanumeric character, then
         add ch to postfix expression
      else if ch = opening parenthesis (, then
         push ( into stack
      else if ch = ^, then            //exponential operator of higher precedence
         push ^ into the stack
      else if ch = closing parenthesis ), then
         while stack is not empty and stack top ≠ (,
            do pop and add item from stack to postfix expression
         done

         pop ( also from the stack
      else
         while stack is not empty AND precedence of ch <= precedence of stack top element, do
            pop and add into postfix expression
         done

         push the newly coming character.
   done

   while the stack contains some remaining characters, do
      pop and add to the postfix expression
   done
   return postfix
End

例子

#include<iostream>
#include<stack>
#include<locale>      //for function isalnum()
using namespace std;

int preced(char ch) {
   if(ch == '+' || ch == '-') {
      return 1;              //Precedence of + or - is 1
   }else if(ch == '*' || ch == '/') {
      return 2;            //Precedence of * or / is 2
   }else if(ch == '^') {
      return 3;            //Precedence of ^ is 3
   }else {
      return 0;
   }
}

string inToPost(string infix ) {
   stack<char> stk;
   stk.push('#');               //add some extra character to avoid underflow
   string postfix = "";         //initially the postfix string is empty
   string::iterator it;

   for(it = infix.begin(); it!=infix.end(); it++) {
      if(isalnum(char(*it)))
         postfix += *it;      //add to postfix when character is letter or number
      else if(*it == '(')
         stk.push('(');
      else if(*it == '^')
         stk.push('^');
      else if(*it == ')') {
         while(stk.top() != '#' && stk.top() != '(') {
            postfix += stk.top(); //store and pop until ( has found
            stk.pop();
         }
         stk.pop();          //remove the '(' from stack
      }else {
         if(preced(*it) > preced(stk.top()))
            stk.push(*it); //push if precedence is high
         else {
            while(stk.top() != '#' && preced(*it) <= preced(stk.top())) {
               postfix += stk.top();        //store and pop until higher precedence is found
               stk.pop();
            }
            stk.push(*it);
         }
      }
   }

   while(stk.top() != '#') {
      postfix += stk.top();        //store and pop until stack is not empty.
      stk.pop();
   }

   return postfix;
}

int main() {
   string infix = "x^y/(5*z)+2";
   cout << "Postfix Form Is: " << inToPost(infix) << endl;
}