题解 | #逆波兰表达式求值#

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 evalRPN(vector<string> &tokens) {
        // write code here
        vector<int> mStack;

        for (int i = 0; i < tokens.size(); i++) {
            string s = tokens[i];
            if (isOp(s)) {
                int b = mStack.back();
                mStack.pop_back();
                int a = mStack.back();
                mStack.pop_back();


                if (s == "+") {
                    mStack.push_back(a + b);
                } else if (s == "-") {
                    mStack.push_back(a - b);
                } else if (s == "*") {
                    mStack.push_back(a * b);
                } else if (s == "/") {
                    mStack.push_back(a / b);
                } else {

                }

            }
            else{
                int num = stoi(s);
                mStack.push_back(num);
            }
        }

        return mStack.back();
    }

  //判断是否是操作符
    bool isOp(string s) {
        if (s == "+" || s == "-" || s == "*" || s == "/")return true;
        return false;
    }
};