堆棧處理語言recognizng字符串，java程序

Question

所以我的計劃是關於語言L = {'w$w' : w是一個字符超過$, w' = reverse(w)}

所以其他可能的空字符串時，像HOD $ DOH被送到isINlanguage函數的參數中，應該返回true，但我的程序只是停止和掛起犯規了放東西

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

你是不是重置ch內while循環到下一個字符：

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

此外，在try塊內，轉換不是必需的。分配：

stackTop = (char) aStack.pop(); 

...是更好的寫法如下：

stackTop = aStack.pop(); 

---

順便說一句，你真的複雜化你的任務，通過使用boolean變量，try-catch塊。不要讓stack.pop()發生任何異常。相反，只有在堆棧不空時才彈出一個元素。此外，只要您發現字符串不匹配所需的語言，就可以直接返回，因此不需要布爾變量。

我會修改你的方法：

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++)將在指數i獲得字符，然後遞增i。