题解 | #简单计算器#

//本题其实是典型的使用栈来计算后缀表达式，照着后缀表达式的计算方式即可
//这里我用map来存储运算符的优先级
//本题还有一个很重要的问题就是一个数字可能有多位数要记得判断
#include <iostream>
#include<stack>
#include<string>
#include <map>
#include<cctype>
using namespace std;

double calculate(double a, double b, char op) {
    double n1 = double(a);
    double n2 = double(b);
    switch (op) {
        case'+':
            return n1 + n2;
            break;
        case '-':
            return n1 - n2;
            break;
        case '*':
            return n1 * n2;
            break;
        case '/':
            return n1 / n2;
            break;

    }
    return 0;
}
//一个数字可能有多位数所以需要往后查看
double GetNumber(string str, int& index) {
    double number = 0;
  //判断后面的书符是否是数字
    while (str[index] >= '0' && str[index] <= '9') {
        number = number * 10 + str[index] - '0';
        index++;
    }
    return number;
}

int main() {
    string str;
    map<char, int> m;//存储运算符优先级
    m['+'] = 1;
    m['-'] = 1;
    m['*'] = 2;
    m['/'] = 2;

    while (getline(cin, str)) {
        if (str == "0") break;
        stack<double> s1;
        stack<char> op;
        for (int i = 0; i < str.size(); i++) {
            if (str[i] >= '0' && str[i] <= '9') {
                s1.push(GetNumber(str, i));
            } else {
                if (str[i] == ' ')
                    continue;
                else {

                    if (op.empty() || m[op.top()] < m[str[i]]) {
                        op.push(str[i]);
                    } else {
                        while (m[op.top()] >= m[str[i]]) {
						  //将运算符栈内所有大于当前遍历运算符的运算全部运算
                            double b = s1.top();
                            char ch = op.top();
                            op.pop();

                            s1.pop();
                            double a = s1.top();
                            s1.pop();
                            s1.push(calculate(a, b, ch));
                            if (op.empty()) break;
                        }
                        op.push(str[i]);

                    }
                }
            }
        }
        while (!s1.empty()) {
            double b = s1.top();
            char ch = op.top();
            op.pop();
            s1.pop();
            double a = s1.top();
            s1.pop();
            s1.push(calculate(a, b, ch));
            if (s1.size() == 1) {
                printf("%0.2f\n", calculate(a, b, ch));
                break;
            }
        }
    }
}