比较Java(回文)中的堆栈弹出和队列出列

时间:2012-06-08 16:14:04

标签: java linked-list palindrome

完全披露:这是针对作业的,所以请不要发布实际的代码解决方案!

我有一个赋值,要求我从用户那里取一个字符串并将其传递给一个堆栈和一个队列,然后使用这两个来比较这些字符以确定该字符串是否为回文。我编写了程序,但似乎在某处出现了一些逻辑错误。这是相关的代码:

public static void main(String[] args) {

    UserInterface ui = new UserInterface();
    Stack stack = new Stack();
    Queue queue = new Queue();
    String cleaned = new String();
    boolean palindrome = true;

    ui.setString("Please give me a palindrome.");
    cleaned = ui.cleanString(ui.getString());

    for (int i = 0; i < cleaned.length(); ++i) {
        stack.push(cleaned.charAt(i));
        queue.enqueue(cleaned.charAt(i));
    }

    while (!stack.isEmpty() && !queue.isEmpty()) {
        if (stack.pop() != queue.dequeue()) {
            palindrome = false;
        }
    }

    if (palindrome) {
        System.out.printf("%s is a palindrome!", ui.getString());
    } else
        System.out.printf("%s is not a palindrome :(", ui.getString());

    stack.dump();
    queue.clear();

}

 public class Stack {

   public void push(char c) {
    c = Character.toUpperCase(c);
    Node oldNode = header;
    header = new Node();
    header.setData(c);
    header.setNext(oldNode);
  }

  public char pop() {
    Node temp = new Node();
    char data;
    if (isEmpty()) {
        System.out.printf("Stack Underflow (pop)\n");
        System.exit(0);
    }
    temp = header;
    data = temp.getData();
    header = header.getNext();
    return data;
  }

}

public class Queue {

  public void enqueue(char c) {
    c = Character.toUpperCase(c);
    Node n = last;
    last = new Node();
    last.setData(c);
    last.setNext(null);
    if (isEmpty()) {
        first = last;
    } else n.setNext(last);     
  }

  public char dequeue() {
    char data;
    data = first.getData();
    first = first.getNext();
    return data;
  }

}

public String cleanString(String s) {
    return s.replaceAll("[^A-Za-z0-9]", "");
}

基本上,当我在Eclipse中通过调试器运行我的代码时,我的pop和dequeue方法似乎只选择某些字母数字。我正在使用replaceAll("[^A-Za-z0-9]", "")来“清理”用户的任何非字母数字字符串(!,?,&amp;等)。当我说它只选择某些字符时,似乎没有任何我能辨别的模式。有什么想法吗?

2 个答案:

答案 0 :(得分:0)

假设您的队列和堆栈正确(我使用jdk中的Deque实现尝试了这一点),您的通用算法可以正常工作。由于你的任务涉及数据结构,我几乎只是采用了你的主逻辑并用ArrayDequeue替换了数据结构,所以我不觉得我正在为你回答这个问题。

    String word = "ooffoo";

    word = word.replaceAll("[^A-Za-z0-9]", "");

    Deque<Character> stack = new ArrayDeque<Character>(word.length());
    Deque<Character> queue = new ArrayDeque<Character>(word.length());

    for (char c : word.toCharArray()) {
        stack.push(c);
        queue.add(c);
    }

    boolean pal = true;

    while (! stack.isEmpty() && pal == true) {
        if (! stack.pop().equals(queue.remove())) {
            pal = false;
        }
    }

    System.out.println(pal);

答案 1 :(得分:0)

我建议使用调试器来查看正在比较的内容,或者至少吐出一些打印行:

while (!stack.isEmpty() && !queue.isEmpty()) {
    Character sc = stack.pop();
    Character qc = queue.dequeue();
    System.out.println(sc + ":" + qc);
    if (sc != qc) {
        palindrome = false;
    }
}