栈-直接计算中缀表达式

 思路与“中缀表达式转后缀表达式”相似

#include <iostream>
#include <cstring>
#include <stack>
#include <cmath>
using namespace std;

stack<double> num;
stack<char> oprt;

char s[1005];
int l = 0;

void in(){
    char t[1005];
    cin.getline(t, 1000);
    int ll = strlen(t);
    for(int i = 0; i < ll; ++i){
        if(t[i] != ' '){
            s[l++] = t[i];
        }
    }
}

double cal(double a, double b, char o){
    if(o == '+') return a+b;
    if(o == '-') return a-b;
    if(o == '*') return a*b;
    if(o == '/') return a/b;
    else{
        cerr << "error" << endl;
    }
}

int main()
{
    in();
    int i = 0;
    while(i < l){
        if(isdigit(s[i]) || s[i] == '.'){
            double b = 0;
            while(i < l && isdigit(s[i])){
                b *= 10;
                b += s[i++] - '0';
            }
            if(s[i] == '.'){
                ++i;
                int ct = 0;
                while(i < l && isdigit(s[i])){
                    b += (s[i++]-'0') * pow(0.1, ++ct);
                }
            }
            num.push(b);
        }
        else{
            switch(s[i]){
            case '+':
            case '-':
                while(!oprt.empty() && oprt.top() != '('){ // 如果用if，那么形如“1+2*3+”的式子就不能正常处理
                    double b = num.top(); num.pop();
                    double a = num.top(); num.pop();
                    num.push(cal(a,b,oprt.top())); oprt.pop();
                }
                oprt.push(s[i]);
                break;
            case '*':
            case '/':
                if(!oprt.empty() && (oprt.top() == '*' || oprt.top() == '/')){
                    double b = num.top(); num.pop();
                    double a = num.top(); num.pop();
                    num.push(cal(a,b,oprt.top())); oprt.pop();
                }
                oprt.push(s[i]);
                break;
            case '(':
                oprt.push(s[i]);
                break;
            case ')':
                while(!oprt.empty() && oprt.top() != '('){
                    double b = num.top(); num.pop();
                    double a = num.top(); num.pop();
                    num.push(cal(a,b,oprt.top())); oprt.pop();
                }
                oprt.pop(); // '('
                break;
            }
            ++i;
        }
    }
    while(!oprt.empty()){
        double b = num.top(); num.pop();
        double a = num.top(); num.pop();
        num.push(cal(a,b,oprt.top())); oprt.pop();
    }
    printf("%.2f\n", num.top());
    system("pause");
    return 0;
}

数月后另解，用运算符优先级确定运算顺序，更为本质：

#include <iostream>
#include <cstring>
#include <algorithm>
#include <stack>
#include <map>
using namespace std;

char s[605];
stack<int> num;
stack<char> oprt;
map<char, int> priority = {{'+', 1}, {'-', 1}, {'*', 2}, {'/', 2}, {'(', 0}, {')', 0}};

int cal(int a, int b, char op) {
    // printf("calculating %d %c %d\n", a, op, b);
    if (op == '+') return a+b;
    if (op == '-') return a-b;
    if (op == '*') return a*b;
    if (b == 0)
        cerr << "0 divisor" << endl;
    if (op == '/') return a/b;
    cerr << "bad op" << endl;
    return 0;
}

void cal_and_push() {
    int b = num.top(); num.pop();
    int a = num.top(); num.pop();
    char op = oprt.top(); oprt.pop();
    int rst = cal(a, b, op);
    num.push(rst);
}

int main()
{
    int t; cin >> t; getchar();
    while (t--) {
        cin.getline(s, 605);
        int i = 0;
        int l = strlen(s);
        while (i < l) {
            if (isdigit(s[i])) {
                int n = 0;
                while (i < l && isdigit(s[i])) {
                    n *= 10;
                    n += s[i] - '0';
                    ++i;
                }
                num.push(n);
            }
            else {
                char c = s[i++];
                if (c == '(') {
                    oprt.push(c);
                }
                else if (c == ')') {
                    while (!oprt.empty() && oprt.top() != '(')
                        cal_and_push();
                    oprt.pop(); // for '('
                }
                else {
                    while (!oprt.empty() && priority[oprt.top()] >= priority[c])
                        cal_and_push();
                    oprt.push(c);
                }
            }
        }
        while (!oprt.empty())
            cal_and_push();
        printf("%d\n", num.top());
        num.pop();
    }
    return 0;
}