题目是这样的，要求实现一个带有min函数的栈

下面是我的代码

public class Solution {

    Stack<Integer> stack = new Stack<>();
    Stack<Integer> minStack = new Stack<>();

    public void push(int node) {
        stack.push(node);
        if (minStack.isEmpty()){
            minStack.push(node);
        }
        if (!minStack.isEmpty() && minStack.peek() > node){

            minStack.push(node);
        }
    }

    public void pop() {
        if (stack.isEmpty() && minStack.isEmpty()){
            return;
        }
        stack.pop();
        if (stack.peek().equals(minStack.peek())){
            minStack.remove(stack.peek());
        }
    }

    public int top() {
        return stack.peek();
    }

    public int min() {
        
        return minStack.peek();
    }

    public static void main(String[] args) {
        Solution solution = new Solution();
        solution.push(3);
        System.out.println(solution.min());
        solution.push(4);
        System.out.println(solution.min());
        solution.push(2);
        System.out.println(solution.min());
        solution.push(3);
        System.out.println(solution.min());
        solution.pop();
        System.out.println(solution.min());
        solution.pop();
        System.out.println(solution.min());
        solution.pop();
        System.out.println(solution.min()); //这行会遇到遇到问题
        solution.push(0);
        System.out.println(solution.min());
    }
}

在提交时会遇到这个问题

如果把代码中出问题的行注释掉就会的到正确的结果，也就是在当minStack为空时就什么都不返回，可是什么都不返回应该怎么写啊。min()方法总是要返回一个值，抛出异常提交也不能通过，

这下应该怎么改啊