思路

利用栈即可解决，主要坑有操作数1和操作数2的区别

代码

class Solution {
public:int evalRPN(vector<string>& tokens) {int res = 0;stack<int> num;for(int i = 0; i < tokens.size(); i++){if(tokens[i] == "+" || tokens[i] == "-" || tokens[i] == "*" || tokens[i] == "/"){int opt1 = num.top();num.pop();int opt2 = num.top();num.pop();if(tokens[i] == "+"){num.push(opt1 + opt2);}else if(tokens[i] == "-"){num.push(opt2 - opt1);}else if(tokens[i] == "*"){num.push(opt1 * opt2); }else if(tokens[i] == "/"){num.push(opt2 / opt1);}}else{int tmp = 0, d = 1;for(int j = tokens[i].size()-1; j >=0; j--){if(j == 0 && tokens[i][0] == '-'){tmp *= -1;break;}tmp += d * (tokens[i][j] - '0');d *= 10;}num.push(tmp);}}return num.top(); }
};