https://leetcode.com/problems/evaluate-reverse-polish-notation/?tab=Description
开一个栈，遇到数字压栈，遇到运算符就从栈中弹出两个元素做相应的计算然后再压回去。
Time complexity: O(n)
Space complexity: O(n)
class Solution { public:int evalRPN(vector<string>& tokens) {if (tokens.empty()) return 0;if (tokens.size() == 1) return stoi(tokens[0]);stack<int> stk;int ret = 0;for (int i = 0; i < tokens.size(); ++i) {string& str = tokens[i];if (isop(str)) {int x = stk.top(); stk.pop();int y = stk.top(); stk.pop();if (str == "+") {ret = y + x;} else if (str == "-") {ret = y - x;} else if (str == "*") {ret = y * x;} else {ret = y / x;}stk.push(ret);} else {stk.push(stoi(str));}}return ret;} private:bool isop(string& str) {return (str == "+" || str == "-" || str == "*" || str == "/");} };