题目描述：

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

思路：遍历逆波兰式的每一项。如果是数字的话，则入栈。如果是运算符，则让栈顶的两个元素出栈，做相应的运算后，将运算的结果入栈。遍历结束后，栈中剩余的元素即为逆波兰式的运算结果。

代码：

int evalRPN(vector<string> &tokens)
{
    int length = tokens.size();
    stack<int> num_stack;
    for(int i = 0;i < length;i++)
    {
        if(strcmp(tokens[i].c_str(),"*") == 0)
        {
            int number1 = num_stack.top();
            num_stack.pop();
            int number2 = num_stack.top();
            num_stack.pop();
            int result = number1 * number2;
            num_stack.push(result);
        }
        else if(strcmp(tokens[i].c_str(),"/") == 0)
        {
            int number1 = num_stack.top();
            num_stack.pop();
            int number2 = num_stack.top();
            num_stack.pop();
            int result = number1 / number2;
            num_stack.push(result);
        }
        else if(strcmp(tokens[i].c_str(),"+") == 0)
        {
            int number1 = num_stack.top();
            num_stack.pop();
            int number2 = num_stack.top();
            num_stack.pop();
            int result = number1 + number2;
            num_stack.push(result);
        }
        else if(strcmp(tokens[i].c_str(),"-") == 0)
        {
            int number1 = num_stack.top();
            num_stack.pop();
            int number2 = num_stack.top();
            num_stack.pop();
            int result = number1 - number2;
            num_stack.push(result);
        }
        else
        {
            int num = atoi(tokens[i].c_str());
            num_stack.push(num);
        }
    }
    return num_stack.top();
}

LeetCode:Evaluate Reverse Polish Notation