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所以我试图在Java中实现Shunting-yard算法，我做得很好，但我的问题是我希望由算法评估的文本字符串包含多于一位的数字。

我有这个问题的解决方案，但解决方案并不是那么“漂亮”。我能做的是，每当找到一个数字时（在主 for 循环中），我会将它连接到一个字符串并将一个布尔变量变为 true，以让程序知道我们在一个数字内。当 for 循环找到一个非数字字符时，布尔变量将变为 false，我们照常继续算法。

我觉得这个解决方案不是很好，因为每当我遇到困难时，将布尔变量 true/false 一直是我的解决方案。

请问有什么帮助吗？

    while there are tokens to be read:
    read a token.
    **if the token is a number**, then push it to the output queue.
    if the token is an operator, then:
        while there is an operator at the top of the operator stack with
            greater than or equal to precedence:
                pop operators from the operator stack, onto the output queue;
        push the read operator onto the operator stack.
    if the token is a left bracket (i.e. "("), then:
        push it onto the operator stack.
    if the token is a right bracket (i.e. ")"), then:
        while the operator at the top of the operator stack is not a left bracket:
            pop operators from the operator stack onto the output queue.
        pop the left bracket from the stack.
        /* if the stack runs out without finding a left bracket, then there are
        mismatched parentheses. */
if there are no more tokens to read:
    while there are still operator tokens on the stack:
        /* if the operator token on the top of the stack is a bracket, then
        there are mismatched parentheses. */
        pop the operator onto the output queue.
exit.

我已经突出显示了算法中出现问题的那一行（第 3 行）。它说数字，但我实现算法的方式只能捕获数字。

谢谢！

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