java - 计算逆波兰表示法中算术表达式的值

Question

我们给出了字符串数组，其中包含后缀表达式作为其元素，我们需要进行评估

我尝试使用堆栈但遇到异常:线程“main”中的异常 java.util.EmptyStackException

public int evalRPN(String[] A) {

     Stack<Integer> s=new Stack<>();
     int i=0;
     while(s.isEmpty()==false && i<A.length)
     {
        String p=A[i++];
        if(p.matches("\\d"))
            s.push(Integer.parseInt(p));
        else
            {
                int a=s.pop();
                int b=s.pop();

                switch(p)
                {
                    case "+": s.push(b+a);break;
                    case "-": s.push(b-a);break;
                    case "*": s.push(b*a);break;
                    case "/": s.push(b/a);break;


                }
            }

    }
    return s.pop();
 }

Answer 1

public int evalRPN(String[] A) {

     Stack<Integer> s=new Stack<>();
     int i=0;
     while(i<A.length) //stack will be empty right at the start.. no need to check this
     {
        String p=A[i++];
        if(p.matches("\\d")) //I am not quite sure, but this just matches single digits, right? I would go for "\\d*", but I am not completely sure. Anyways you won't capture negative values or decimals.. perhaps better exclude +-*/ and use Double.parseDouble and a Stack<Double>
            s.push(Integer.parseInt(p));
        else
            {
                int a=s.pop();
                int b=s.pop();

                switch(p)
                {
                    case "+": s.push(b+a);break;
                    case "-": s.push(b-a);break;
                    case "*": s.push(b*a);break;
                    case "/": s.push(b/a);break;


                }
            }

    }
    return s.pop();
 }

如果给定的表达式不正确(运算符太多/操作数太少)，这可能会遇到错误，或者可能只返回中间结果(操作数太多/运算符太少)。