150. Evaluate Reverse Polish Notation(逆波兰式)

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题目链接：https://leetcode.com/problems/evaluate-reverse-polish-notation/

题意：

Evaluate the value of an arithmetic expression in Reverse Polish Notation.

Valid operators are +, -, *, /. Each operand may be an integer or another expression.

Some examples:

  ["2", "1", "+", "3", "*"] -> ((2 + 1) * 3) -> 9  ["4", "13", "5", "/", "+"] -> (4 + (13 / 5)) -> 6

此题考察点：逆波兰式，后缀表达式。借用栈可以解决。此题另外用到了C++11中新加的几个库函数。to_string函数，这是C++11新增的，使用非常方便，简单查了下：C++11标准增加了全局函数std::to_string，以及std::stoi/stol/stoll等等函数（这几个就是string转int，long，以及long long啦~）
to_string这个函数还是很强大的！
string to_string (int val);string to_string (long val);string to_string (long long val);string to_string (unsigned val);string to_string (unsigned long val);string to_string (unsigned long long val);string to_string (float val);string to_string (double val);string to_string (long double val)
class Solution {public:    int evalRPN(vector<string>& tokens) {        stack<string> s;        for(auto token:tokens){            if(!is_operator(token)){                s.push(token);            }            else{                int y = stoi(s.top());                s.pop();                int x = stoi(s.top());                s.pop();                if(token[0]=='+') x+=y;                else if(token[0]=='-') x-=y;                else if(token[0]=='*') x*=y;                else x/=y;                s.push(to_string(x));            }        }        return stoi(s.top());    }private:    bool is_operator(const string &op){        return op.size()==1 && string("+-*/").find(op)!= string::npos;    }};

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