150. Evaluate Reverse Polish Notation(逆波兰式)

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;
}
};