java - 自定义列表类，如何循环hasNext方法？

Question

大家好，我是 java 新手，所以我非常感谢对此的任何帮助。好吧，这就是问题所在:我有一个列表类和一个 listNode 类，列表类由名称、firstNode 和lastNode 表示。firstNode 和 lastNode 来自类型 listNode，listNode 由一个对象(例如 data 或 Object o)和一个 nextNode 表示，后者指向列表中的下一个节点，该节点也来自类型 listNode。

列表类:

public class List {

private ListNode firstNode;
private ListNode lastNode;
private String name;

public List() {
    this("list");
}

public List(String listName) {
    name = listName;
    firstNode = lastNode = null;
}

public void insertAtFront(Object insertItem) {
    if (isEmpty())
        firstNode = lastNode = new ListNode(insertItem);
    else
        firstNode = new ListNode(insertItem, firstNode);
}

public void insertAtBack(Object insertItem) {
    if (isEmpty())
        firstNode = lastNode = new ListNode(insertItem);
    else
        lastNode = lastNode.nextNode = new ListNode(insertItem);
}

public Object removeFromFront() throws EmptyListException {
    if (isEmpty())
        throw new EmptyListException(name);
    Object removedItem = firstNode.data;

    if (firstNode == lastNode)
        firstNode = lastNode = null;
    else
        firstNode = firstNode.nextNode;
    return removedItem;
}

public Object removeFromBack() throws EmptyListException {
    if (isEmpty())
        throw new EmptyListException(name);

    Object removedItem = lastNode.data;
    if (firstNode == lastNode)
        firstNode = lastNode = null;
    else {
        ListNode current = firstNode;

        while (current.nextNode != lastNode)
            current = current.nextNode;

        lastNode = current;
        current.nextNode = null;
    }
    return removedItem;
}

public boolean isEmpty() {
    return firstNode == null;
}

public void print() {
    if (isEmpty()) {
        System.out.printf("Empty %s\n", name);
        return;
    }
    System.out.printf("The %s is : ", name);
    ListNode current = firstNode;

    while (current != null) {
        System.out.printf("%s", current.data);
        current = current.nextNode;
    }
    System.out.println("\n");
}

@Override
public String toString() {
    String stk = "(";
    if(isEmpty())return "Empty List";
    ListNode checkNode = firstNode;
        while (checkNode != null) {
        stk += checkNode.data.toString()+ " , ";
        checkNode = checkNode.nextNode;
    }
    return stk+")";
}
public ListNode removeAt (int k){
    if(k<=0 || k>getLength())
        try{
            throw new IllegalValues();
        }catch(IllegalValues iv){
            iv.printStackTrace();
            return null;
        }
    ListNode newNode = firstNode;
    if(k==1){
        newNode = firstNode;
        firstNode = firstNode.nextNode;
        return  newNode;
    }
    if(k==2){
        newNode = firstNode.nextNode;
        firstNode.nextNode = firstNode.nextNode.nextNode;
        return newNode;
    }

    if(k==3){
        newNode = firstNode.nextNode;
        firstNode.nextNode.nextNode = firstNode.nextNode.nextNode.nextNode;
        return newNode;
    }

    if(k==4){
        newNode = firstNode.nextNode;
        firstNode.nextNode.nextNode.nextNode = firstNode.nextNode.nextNode.nextNode.nextNode;
        return newNode;
    }

    return newNode;
}
public int getLength(){
    ListNode checkNode = firstNode;
    int count =0;
    while (checkNode != null) {
    count++;
    checkNode = checkNode.nextNode;
}
    return count;
}

}

列表节点类:

public class ListNode {

Object data;
ListNode nextNode;

public ListNode(Object o) {
    this(o, null);
}

public ListNode(Object o, ListNode node) {
    data = o;
    nextNode = node;
}

public Object getObject() {
    return data;
}

public ListNode getNext(){
    return nextNode;
}

}

这是我正在使用的两个类。 我的问题是removeAt()方法我不知道如何概括它并为所有代码(如for语句)做出一般性答案我只能通过在if语句中单独编写每个案例来使其工作。我需要编写一个 for 循环，以某种方式可以循环抛出 hasNext() 方法。有任何想法吗？？提前致谢

Answer 1

好吧，为了删除第 k 个节点(当 k>1 时)，您始终应该执行一个节点:

someNode.nextNode = someNode.nextNode.nextNode;

这是您要删除的节点之前的节点，这使其成为第 k-1 个节点。

您可以通过 for 循环找到该节点:

if (k==1) {
    ListNode removedNode = firstNode;
    firstNode = firstNode.nextNode;
    return removedNode;
}
ListNode someNode = firstNode;
for (int i = 1; i < k - 1; i++) {
    someNode = someNode.nextNode;
}
ListNode removedNode = someNode.nextNode;
someNode.nextNode = someNode.nextNode.nextNode;
return removedNode;

请注意，k==1 的情况是单独处理的。