携程 2021.3.4 笔试题

class Solution {
public:
    int evalRPN(vector<string>& tokens) {
        stack<int> stk;

        for (int i = 0; i < tokens.size(); i ++ )
        {
            if (tokens[i] == "+" || tokens[i] == "-" || tokens[i] == "*" || tokens[i] == "/")
            {
                int a = stk.top();                      //取得栈顶的两个数
                stk.pop();
                int b = stk.top();
                stk.pop();

                if (tokens[i] == "+") stk.push(b + a);  //计算结果，再推入栈，注意这里a，b要反过来
                if (tokens[i] == "-") stk.push(b - a);
                if (tokens[i] == "*") stk.push(b * a);
                if (tokens[i] == "/") stk.push(b / a);
            }                                   //当前为数字就直接入栈，为运算符就出栈两个数来计算
            else stk.push(stoi(tokens[i]));     //stoi表示string to int，字符串转整数
        }

        int res = stk.top();                    //最后剩下的栈顶元素即是逆波兰表达式的计算结果

        return res;
    }
};