删除并返回索引链接列表中的项目

Question

package linkedList.list;

import linkedList.node.ListNode;

public interface LinkedList<N extends ListNode<T>,T>
{
   
    public boolean isEmpty();

    
    public int size();

   
    public String toString();

    
    public T[] toArray(Class<? extends T> cl);

    
    public LinkedList<N,T> fromArray(T[] array) throws ListAccessError;
}

package linkedList.list;

import linkedList.node.ListNode;

import java.lang.reflect.Array;
import java.util.ArrayList;
import java.util.Arrays;

public abstract class BasicList<N extends ListNode<T>,T> implements LinkedList<N,T> {

    N root;

    int size;

    public boolean isEmpty() {
        return size == 0;
    }

    public int size() {
        return size;
    }

    public N getRoot() {
        return root;
    }

   
    public void setRoot(N newRoot) {
        root = newRoot;
    }

   
    public T[] toArray(Class<? extends T> cl) {
        T[] array = (T[]) Array.newInstance(cl,size());
        ListNode<T> node = getRoot();
        for (int index = 0; index < size(); index++) {
            array[index] = node.getValue();
            node = node.getNext();
        }
        return array;
    }

   
    public String toString() {
        if (isEmpty()) {
            return "[]";
        } else {
            ListNode<T> currentNode = getRoot();
            StringBuilder string = new StringBuilder("[" + currentNode.getValue());
            while ((currentNode = currentNode.getNext()) != null) {
                string.append("," + currentNode.getValue());
            }
            string.append("]");
            return string.toString();
        }
    }

   
    public String toString(int n) {
        if (isEmpty()) {
            return "[]";
        } else {
            ListNode<T> currentNode = getRoot();
            StringBuilder string = new StringBuilder("[" + currentNode.getValue());
            int added = 0;
            while (added < n && (currentNode = currentNode.getNext()) != null) {
                string.append("," + currentNode.getValue());
                added++;
            }
            if (currentNode != null) {
                string.append(",...");
            }
            string.append("]");
            return string.toString();
        }
    }
}

package linkedList.list;

import com.sun.xml.internal.bind.v2.runtime.unmarshaller.XsiNilLoader;
import jdk.nashorn.internal.ir.IfNode;
import linkedList.node.ListNode;
import linkedList.node.SingleLinkNode;

import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;

public class SingleLinkList<T> extends BasicList<SingleLinkNode<T>,T> implements List<T> {

    public SingleLinkList<T> fromArray(T[] array) throws ListAccessError {
        for (int index = array.length-1; index >= 0; index--) {
            add(0,array[index]);
        }
        return this;
    }

    ListNode<T> getNode(int index) throws ListAccessError {
        // Is the list empty?  If so, cannot access the node.
        if (isEmpty()) {
            throw new ListAccessError("Cannot get node.  List is empty.");
        }
        // Is the given index negative?  If so, this is an error.
        if (index < 0) {
            throw new ListAccessError("Cannot get node.  Negative index.");
        }

        ListNode<T> currentNode = getRoot(); // start at the root
        while (index != 0 && currentNode != null) { // walk along the list (if haven't reached the end by hitting null node)
            currentNode = currentNode.getNext(); // by gettting next node in the list
            index--; // and reducing index by one
        }
        // Reached the end of the list (by hitting null node)?  If so, cannot access the required node.
        if (currentNode == null) {
            throw new ListAccessError("Cannot get node.  Not enough nodes in the list.");
        }
        // Successfully found node by walking through until index was zero.
        return currentNode;
    }

    public T get(int index) throws ListAccessError {
        return getNode(index).getValue();
    }

    @Override
    public void add(int index, T value) throws ListAccessError {
        if (index > size() || index < 0) {
            throw new ListAccessError("Index bigger than size.");
        }

        SingleLinkNode<T> newNode = new SingleLinkNode<T>(value);
        SingleLinkNode<T> current = getRoot();

        if(index==0) {
            setRoot(newNode);
            newNode.setNext(current);
            size++;
        } else {
            while(--index > 0) {
                current = current.getNext();
            }
            newNode.setNext(current.getNext());
            current.setNext(newNode);
            size++;
        }
    }

    @Override
    public T remove(int index) throws ListAccessError {
        if (index >= size() || index < 0) {
            throw new ListAccessError("Index out of bounds");
        }
        if (isEmpty()) {
            throw new ListAccessError("List is empty cannot remove from list");
        }

        SingleLinkNode<T> current = getRoot();
        SingleLinkNode<T> nextItem = getRoot().getNext();
        

        return null;
    }
}
package linkedList.node;

public class SingleLinkNode<T> implements ListNode<T>
{

  private T value;

  private SingleLinkNode<T> next;

  public SingleLinkNode(T value) {
    this.value = value;
    next = null;
  }

  public SingleLinkNode(T value, SingleLinkNode<T> next) {
    this.value = value;
    this.next = next;
  }

  public T getValue() {
    return value;
  }

  @Override
  public void setValue(T value) {
    this.value = value;
  }

  public SingleLinkNode<T> getNext() {
    return next;
  }

  @Override
  public void setNext(ListNode<T> node) {
    next = (SingleLinkNode<T>) node;
  }

  public String toString() {
    return value + (getNext() != null ? "=>" + getNext() : "");
  }
}

我刚刚实现了 add 方法，它获取索引和值并将其插入到列表中的那个位置。但是，在研究如何在索引处删除时，我无法找到仅使用给定索引删除节点的任何内容。欢迎任何帮助，并提前感谢任何解决方案！

Answer 1

首先，您必须检查这是否是您必须删除的第 0 个索引，因此您必须转移根目录。然后要删除第 i 个根的节点，您所要做的就是：

nodeBeforeI.setNext(nodeBeforeI.getNext().getNext()); 

完整代码：

public T remove(int index) throws ListAccessError {
        if (index >= size() || index < 0) {
            throw new ListAccessError("Index out of bounds");
        }
        if (isEmpty()) {
            throw new ListAccessError("List is empty cannot remove from list");
        }

        SingleLinkNode<T> current = getRoot();
        SingleLinkNode<T> removed;
        if (index == 0){
            removed = current;
            setRoot(current.getNext()); // setting root to root.next
        }else{
            for (int i = 0; i < index -1; i++){
                current = current.getNext();
            }
            removed = current.getNext();
            current.setNext(current.getNext().getNext()); 
        }
        

        return removed.getValue();
    }

Answer 2

当从链表中删除 I 位置的节点时，您需要做的就是

1. 到达 i-1 处的节点，我们将其命名为 'n1'
2. 获取节点n2=n1.next.next
3. 使 n1.next =n2

您只需在 i 处删除到所请求节点的“链接”，它就不再是链表的一部分。