Java-我正在将堆栈链表修改为队列链表,但我的出列方法的行为仍然类似于pop
Java-我正在将堆栈链表修改为队列链表,但我的出列方法的行为仍然类似于pop,java,linked-list,stack,queue,Java,Linked List,Stack,Queue,private Node back尚未使用,并且除了重命名某些内容外,enqueue(推送)和dequeue(弹出)尚未真正修改。同样,这最初是一个堆栈,但我正试图将其修改为一个队列。我以前用ints做过非链表队列和堆栈,但是用对象和链表我有点迷路了 public class DogQueue { private Node front = null; private Node back = null; private Node element = null; p
private Node back
尚未使用,并且除了重命名某些内容外,enqueue
(推送)和dequeue
(弹出)尚未真正修改。同样,这最初是一个堆栈,但我正试图将其修改为一个队列。我以前用int
s做过非链表队列和堆栈,但是用对象和链表我有点迷路了
public class DogQueue
{
private Node front = null;
private Node back = null;
private Node element = null;
private int counter = 0;
以上只是设置变量
private class Node //This sets up the Linked List
//Data Structure with nodes.
{
private Dog doggy;
private Node nextNode;
private Node firstNode;
Node(Dog newDog)
{
doggy = newDog;
}
}
上面是我不太理解的节点内容
public void enqueue(Dog aDog) //This should enqueue
//an object of type Dog.
{
Node dogNode = new Node(aDog);
dogNode.nextNode = front;
counter++;
front = dogNode;
}
这里的上述内容未经push方法修改,只是重命名了
public Dog dequeue() //This should output
//the first entry in the list.
{
Dog firstDog = front.doggy;
element = front.firstNode;
counter--;
return firstDog;
}
以上是我遇到最多问题的地方——目前它的行为类似于pop(获取和删除列表中最后输入的元素)
我将计数器设置为刚好上升到5,这样我就可以调试isFull了
public boolean isEmpty() //Checks to see if List is empty
{
if ( counter == 0 )
{
return true;
} else {
return false;
}
}
这只是说,若计数器为零,则isEmpty为真(否则为假)
我不擅长数据结构,但我相信你的排队和出列行为仍然像pop和push一样。 前面应该指向队列的头部,后面应该指向最后一个有效对象。因此尾部最终应该指向null。。 我认为应该是这样的:
public void enqueue(Dog aDog)
{
Node dogNode = new Node(aDog);
counter++;
if (front == null)
front = back = dogNode;
else
{
back.nextNode = dogNode;
back = dogNode;
}
}
public Node dequeue()
{
if(front == null) return null;
Dog firstDog = front ;
front = front.nextNode;
counter--;
return firstDog;
}
public void enqueue(Dog aDog) {
if(front == null) {
front = new Node(aDog);
back = front; // back will move later
} else {
Node tmp = new Node(aDog);
tmp.setFirstNode(back);
back.setNextNode(tmp);
back = tmp;
}
}
public Dog dequeue() {
if(head == null) {
return null;
} else {
Dog tmp = front.getDoggy()
front = front.getNextNode(); //move the front to point to the next location
front.getFirstNode().setNextNode(null); //sever the link to the first element
front.setFirstNode(null); //sever the link to the first element
return tmp;
}
}
这是主要问题。队列是FIFO(先进先出),堆栈是LIFO(后进先出)。对于队列,您排队的第一个元素是您接收的第一个元素,最近推到堆栈上的元素是您接收的第一个元素 为此,让我们稍微检查一下您的代码
public void enqueue(Dog aDog) { //This should enqueue an object of type Dog.
Node dogNode = new Node(aDog);
dogNode.nextNode = front;
counter++;
front = dogNode;
}
您正在将新dog元素的下一个节点设置到前面。您必须转到队列的末尾,将最近的节点设置为新节点,将新节点设置为null。使用您的代码,它看起来像这样:
public void enqueue(Dog aDog)
{
Node dogNode = new Node(aDog);
counter++;
if (front == null)
front = back = dogNode;
else
{
back.nextNode = dogNode;
back = dogNode;
}
}
public Node dequeue()
{
if(front == null) return null;
Dog firstDog = front ;
front = front.nextNode;
counter--;
return firstDog;
}
public void enqueue(Dog aDog) {
if(front == null) {
front = new Node(aDog);
back = front; // back will move later
} else {
Node tmp = new Node(aDog);
tmp.setFirstNode(back);
back.setNextNode(tmp);
back = tmp;
}
}
public Dog dequeue() {
if(head == null) {
return null;
} else {
Dog tmp = front.getDoggy()
front = front.getNextNode(); //move the front to point to the next location
front.getFirstNode().setNextNode(null); //sever the link to the first element
front.setFirstNode(null); //sever the link to the first element
return tmp;
}
}
至少,这确实显示了队列中的第一件事。但它实际上并没有移动头部指针!使用您的代码,要做到这一点,它将如下所示:
public void enqueue(Dog aDog)
{
Node dogNode = new Node(aDog);
counter++;
if (front == null)
front = back = dogNode;
else
{
back.nextNode = dogNode;
back = dogNode;
}
}
public Node dequeue()
{
if(front == null) return null;
Dog firstDog = front ;
front = front.nextNode;
counter--;
return firstDog;
}
public void enqueue(Dog aDog) {
if(front == null) {
front = new Node(aDog);
back = front; // back will move later
} else {
Node tmp = new Node(aDog);
tmp.setFirstNode(back);
back.setNextNode(tmp);
back = tmp;
}
}
public Dog dequeue() {
if(head == null) {
return null;
} else {
Dog tmp = front.getDoggy()
front = front.getNextNode(); //move the front to point to the next location
front.getFirstNode().setNextNode(null); //sever the link to the first element
front.setFirstNode(null); //sever the link to the first element
return tmp;
}
}
你是对的,我以为push的行为像排队一样,但我错了。因此,是的,它们的行为就像在堆栈中一样。编辑:这段代码正确地退出队列,但当我再次尝试退出队列时,它返回一个错误。另外,非常感谢你的帮助。是的,我想你必须检查空值。我忘了。将更新答案。您的返回声明中缺少某些内容。不幸的是,我在链表中迷失了方向,无法填写。我需要花一些时间研究代码,以弄清楚到底发生了什么。return null:D我希望它能起作用:)我也讨厌数据结构:(嗯,在你发帖子逗我笑之前,我试过返回null,但它仍然不能让我第二次排队。另外,我认为“front”应该是“front.doggy”。非常感谢您的帮助,Makoto。对于我的第一篇帖子,StackOverflow非常慷慨。再次感谢。我将试用您的代码并返回。我需要在Node中定义方法getNextNode、get firstNode、setFirstNode和setNextNode,对吗?@instago:是的,您必须定义我使用的帮助器方法。它们遵循f行业中使用的吸气剂和沉降剂。