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
解题思路很清晰，就是通过栈结构来存储操作数，对字符数组进行遍历。当为操作数，则将其入栈，如果为操作符，则弹出栈中的前两个操作数进行相应的算术运算，再将结果入栈。
代码如下：
class Solution { public: int evalRPN(vector<string>& tokens) { stack<int> opera; int sum; for(int i=0;i<tokens.size();i++) { if(tokens[i] == "+" || tokens[i] == "-" || tokens[i] == "*" || tokens[i] == "/") { int one = opera.top(); opera.pop(); int two = opera.top(); opera.pop(); if(tokens[i] == "+") { opera.push(one+two); } else if(tokens[i] == "-") { opera.push(two-one); } else if(tokens[i] == "*") { opera.push(one*two); } else if(tokens[i] == "/") { opera.push(two/one); } } else { opera.push(atoi(tokens[i].c_str())); } } return opera.top(); } };
