LeetCode OJ - Evaluate Reverse Polish Notation

题目：　　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：
　　遇到数字则将其压入栈中，遇到运算符就弹出栈顶的两个元素做相应的运算，然后将运算结果压入栈中。
　　最后的结果保存在栈顶。
代码：

class Solution {
public:
int evalRPN(vector<string> &tokens) {
stack<int> st;
int operand_a = 0, operand_b = 0, tmp;
stringstream ss;
vector<string>::iterator it;
for(it = tokens.begin(); it != tokens.end(); it++){
ss.clear();
if ((*it) == "+" ){
operand_b = st.top(); st.pop();
operand_a = st.top(); st.pop();
st.push(operand_a + operand_b);
}
else if ((*it) == "-" ){
operand_b = st.top(); st.pop();
operand_a = st.top(); st.pop();
st.push(operand_a - operand_b);
}
else if ((*it) == "*" ){
operand_b = st.top(); st.pop();
operand_a = st.top(); st.pop();
st.push(operand_a * operand_b);
}
else if ((*it) == "/" ){
operand_b = st.top(); st.pop();
operand_a = st.top(); st.pop();
st.push(operand_a / operand_b);
}
else{
ss.str(*it);
ss >> tmp;
st.push(tmp);
}
}
return st.top();
}
};