LeetCode OJ

  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();
    }
};