关于逆波兰式的c++实现

正常的表达式 逆波兰表达式

a+b ---> a,b,+

a+(b-c) ---> a,b,c,-,+

a+(b-c)*d ---> a,b,c,-,d,*,+

a+d*(b-c)--->a,d,b,c,-,*,+

a=1+3 ---> a=1,3 +

 

代码运算如下:

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#include "iostream" #include "string" #include "stack" using namespace std;   int main() {     string str;     stack<int> sk;     int s = 0, l = 0, r = 0;     cout << "请输入逆波兰公式:" << endl;     while (cin>>str)     {         if (str[0] == '#')         {             break;         }                   //如果第一个是0-9数字则转换为数字压栈         else if (isdigit(str[0]))         {             sk.push(atoi(str.c_str()));         }         else         {             l = sk.top();             sk.pop();             r = sk.top();             sk.pop();             switch (str[0])             {             case '+':                 s = r + l;                 break;             case '-':                 s = r - l;                 break;             case '*':                 s = r * l;                 break;             case '/':                 s = r / l;                 break;             }             //把计算的结果再次压栈             sk.push(s);         }               }     cout << "结果为:" << s << endl;     system("pause");     return 0; }