（算法）求表达式（不含括号）的值

题目：

求字符串表达式（不含括号）的值

思路：

数字分离

运算符优先级+-*/

只需一个字符串栈，保存数字和运算符

代码：

#include<iostream>
#include<stack>
#include<sstream>
#include<string>
using namespace std;

template<class out_T,class in_T>
out_T convert(const in_T &t){
    stringstream ss;
    out_T result;
    ss<<t;
    ss>>result;
    return result;
}

double calc(string aa,string op,string bb){
    double a=convert<double>(aa);
    double b=convert<double>(bb);
    if(op=="+")
        return a+b;
    if(op=="-")
        return a-b;
    if(op=="*")
        return a*b;
    if(op=="/")
        return a/b;
}

bool isdigit(char s){
    int a=s-'0';
    if(a>=0 && a<=9)
        return true;
    return false;
}

int main(){
    string s;
    string x,y;
    string op;
    double tmp;
    int priv[300];
    priv['+']=priv['-']=2;
    priv['*']=priv['/']=1;
//    priv['(']=10;

    while(cin>>s){
        stack<string> num;
        int len=s.size();
        int start=0;    
        string str;
        char last=0;

        for(int i=0;i<len;i++){
            if(isdigit(s[i])){
                start=i;
                for(;i+1<len && (isdigit(s[i+1]) || s[i+1]=='.');i++);
                str=s.substr(start,i-start+1);
                num.push(str);
            }
            else if(s[i]=='-' && (last==0 || last=='(')){
                num.push("0.0");
                num.push("-");
            }
            else if(priv[s[i]]>0){
                while(priv[s[i]]==2 && num.size()>2){
                    y=num.top();
                    num.pop();
                    op=num.top();
                    num.pop();
                    x=num.top();
                    num.pop();
                    tmp=calc(x,op,y);
                    num.push(convert<string>(tmp));
                }
                num.push(convert<string>(s[i]));
            }
            else
                continue;
            last=s[i];
        }

        while(num.size()>2){
            x=num.top();
            num.pop();
            op=num.top();
            num.pop();
            y=num.top();
            num.pop();
            if(op=="-" || op=="/")
                tmp=calc(y,op,x);
            else
                tmp=calc(x,op,y);
            num.push(convert<string>(tmp));
        }
        cout<<convert<double>(num.top())<<endl;
    }
    return 0;
}