面试题42：计算逆波兰表达式（RPN）

　　这是一个比较简单的题目，借助栈可以轻松实现逆波兰表达式。

题目描述：

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

class Solution {
public:
    int evalRPN(vector<string> &tokens) {
        stack<int> s;
        for(string str : tokens){
            if(str == "+" || str == "-" || str == "*" || str == "/"){
                int a = s.top();s.pop();
                int b = s.top();s.pop();
                if(str == "+"){
                    s.push(b+a);
                }else if(str == "-"){
                    s.push(b - a);
                }else if(str == "*"){
                    s.push(b*a);
                }else{
                    s.push(b/a);
                }
            }else{
                s.push(std::stoi(str));
            }
        }
        return s.top();
    }
};