java - 处理语言、java程序中识别字符串的堆栈

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所以我的程序是关于语言的 L = {'w$w' : w可能是 $, w' = reverse(w)} 之外的空字符串

因此，当像 hod$doh 这样的东西被输入 isINlanguage 函数的参数时，它应该返回 true，但我的程序只是停止并挂起，不会输出任何内容

 import java.util.Stack;


 public class Stacks 
{


public static void main(String[] args){
boolean eval = isInLanguage("sod$dos");

System.out.println(eval);


}




static //   astack.createStack();
    boolean isInLanguage(String aString){
    Stack<Character> aStack = new Stack<>(); 


    int i = 0;
    char ch = aString.charAt(i);
    while (ch != '$') {
        aStack.push(ch);
        i++;
    }
    //Skip the $
    ++i;

    // match the reverse of w
    boolean inLanguage = true; // assume string is in language
    while (inLanguage && i < aString.length()) {
        char stackTop;
        ch = aString.charAt(i);;
        try {
            stackTop =  (char) aStack.pop();
            if (stackTop == ch) {
                i++;
            } else {
                // top of stack is not ch(Charecter do not match)
                inLanguage = false; // reject string

            }
        } catch (StackException e) {
            // aStack.poo() failed, astack is empty (first, half of Stirng
            // is short than second half)

            inLanguage = false;
        }
    }

    if (inLanguage && aStack.isEmpty()) {
        return true;
    }
    else{
        return false;

    }
}
}

Answer 1

您没有将 while 循环内的 ch 重置为下一个字符:

while (ch != '$') {
    aStack.push(ch);
    i++;
    ch = aString.charAt(i);   // Add this
}

此外，在 try block 内，不需要进行强制转换。作业:

stackTop =  (char) aStack.pop();  

...最好写成:

stackTop = aStack.pop();

<小时/>

顺便说一句，通过使用 boolean 变量和 try-catch block ，您的任务确实变得复杂了。不要让 stack.pop() 抛出任何异常。相反，仅当堆栈不为空时才弹出元素。而且，一旦发现字符串与所需的语言不匹配，就可以直接返回，因此不需要 boolean 变量。

我会将你的方法修改为:

static boolean isInLanguage(String aString){
    Stack<Character> aStack = new Stack<>(); 

    int i = 0;
    char ch;

    // This is simplified way to write your first while loop
    // Read a character, move the index further, and test, all in single statement
    while ((ch = aString.charAt(i++)) != '$') {
        aStack.push(ch);
    }

    // Iterate till the stack is not empty
    while (!aStack.isEmpty()) {
       // Get next character in string, and pop an element from stack
       // If they are not equal, return false
        if (aString.charAt(i++) != aStack.pop()) {
            return false;
        }
    }

    // If we reach here, means stack is empty. Test if the index `i` has reached end of the string.
    // If it reached the end of the string, return true, else return false (because there are still some characters to be processed).
    return i == aString.length();
}

aString.charAt(i++) 将获取索引 i 处的字符，并在其后递增 i。