题解 | #牛的表达式计算器#

#include <string>
class Solution {
public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param tokens string字符串vector 
     * @return int整型
     */
    int strToInt(string & str) {
        int sum = 0; 
        bool flag = false;
        if (str[0] == '-') {
            flag = true;
        }
        for (int i = 0; i < str.size(); i++) {
            if (str[i] == '-' || str[i] == '+') continue;
            sum *= 10;
            sum += (str[i]-'0');
        }
        return flag ? -sum: sum;
    }

    int opet(int a, int b, string& op) {
        if (op == "+") {
            return a + b;
        } else if (op == "-") {
            return a - b;
        } else if (op == "*") {
            return a * b;
        } else if (op == "/") {
            return a / b;
        }
        return 0;
    }

    int calculatePostfix(vector<string>& tokens) {
        // write code here
        stack<int> stk;
        int a, b;
        for (int i = 0; i < tokens.size(); i++) {
            if (tokens[i] == "+" || tokens[i] == "-" || tokens[i] == "*" || tokens[i] == "/") {
                b = stk.top();
                stk.pop();
                a = stk.top();
                stk.pop();
                stk.push(opet(a, b, tokens[i]));
            } else {
                stk.push(strToInt(tokens[i]));
            }
        }
        return stk.top();
    }
};