[LeetCode] Evaluate Reverse Polish Notation, Solution

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

[Thoughts]

对于逆波兰式，一般都是用栈来处理，依次处理字符串，

如果是数值，则push到栈里面

如果是操作符，则从栈中pop出来两个元素，计算出值以后，再push到栈里面，

则最后栈里面剩下的元素即为所求。

[code]1:       int evalRPN(vector<string> &tokens) {
2:            stack<int> operand;
3:            for(int i =0; i< tokens.size(); i++)
4:            {
5:                 if ((tokens[i][0] == '-' && tokens[i].size()>1) //negative number
6:                           || (tokens[i][0] >= '0' && tokens[i][0] <= '9')) //positive number
7:                 {
8:                      operand.push(atoi(tokens[i].c_str()));
9:                      continue;
10:                 }
11:                 int op1 = operand.top();
12:                 operand.pop();
13:                 int op2 = operand.top();
14:                 operand.pop();
15:                 if(tokens[i] == "+") operand.push(op2+op1);
16:                 if(tokens[i] == "-") operand.push(op2-op1);
17:                 if(tokens[i] == "*") operand.push(op2*op1);
18:                 if(tokens[i] == "/") operand.push(op2/op1);
19:            }
20:            return operand.top();
21:       }