150. 逆波兰表达式求值——栈的运用

class Solution {
public:
    int evalRPN(vector& tokens) {
        stack nums;
        for(auto token : tokens){
            if(token[0] >= '0' && token[0] <= '9' || token.size() > 1)  //特殊情况是负数，负数的长度必大于1
                nums.push(stoi(token)); //转换为int
            else{
                int num2 = nums.top();  //先出的是后进的
                nums.pop();
                int num1 = nums.top();
                nums.pop();
                switch (token[0])   //剩下就是算术运算符
                {
                case '+':
                    nums.push(num1 + num2);
                    break;
                case '-':
                    nums.push(num1 - num2);
                    break;
                case '*':
                    nums.push(num1 * num2);
                    break;
                case '/':
                    nums.push(num1 / num2);
                    break;
                default:
                    break;
                }
            }
        }
        return nums.top();
    }
};

Accepted
20/20 cases passed (4 ms)
Your runtime beats 98.3 % of cpp submissions
Your memory usage beats 77.45 % of cpp submissions (11.6 MB)