Question

我正在尝试使用java实现链表数据结构，它可以很好地插入或删除第一个元素但是无法通过使用removeLast（）方法删除最后一个元素。

我的链接列表节点类：公共类LLNode { 字符串值; LLNode next;

    public LLNode(String value){
        this.value = value;
        this.next = null;
    }

}

包含头节点的My Linked List类：

public class LL {
    LLNode head;

    public LL(){
        this.head = null;   
    }

    public void insertHead(LLNode input){
        input.next = head;
        this.head = input;
    }

    public void removeFirst(){
        this.head = this.head.next;
    }

    public void removeLast(){
        if (this.head == null || this.head.next == null){
            this.head = null;
            return;
        }
        LLNode current = this.head;
        LLNode tmphead = current;
        while(current.next.next != null){
            current = current.next;
        }
        current.next.next = null;
        this.head = tmphead ;
    }

    public void printAll(){
        LLNode current = this.head;

        while(current != null){
            System.out.print(current.value+" ");
            current = current.next;
        }
        System.out.println();
    }


    public static void main( String[] args){

        LL test = new LL(); 
        LL test2 = new LL();
        String[] eben = {"one","two","three","four","five","six"};

        for(int i =0;i<eben.length;i++){
            test.insertHead(new LLNode(eben[i]));
        }
        test.printAll();
        test.removeFirst();
        test.printAll();
        test.removeLast();
        test.printAll();


    }

}

输出是这样的：

six five four three two one 
five four three two one 
five four three two one

甚至它必须是这样的：

six five four three two one 
five four three two one 
five four three two

我的实施有什么问题？

Answer 1

如果你需要很多代码和/或很多变量来完成这样的任务，你可能会把自己简化为一个角落。我将如何做removeLast()：

public void removeLast() {
  // avoid special case: List is empty
  if (this.head != null) {
    if (this.head.next == null) {
      // handle special case: List has only 1 node
      this.head = null;
    } else {
      LLNode prelast = this.head; // points at node before last (eventually)
      while (prelast.next.next != null) prelast = prelast.next;
      prelast.next = null;
    }
  }
}

未经测试！使用风险由您自行承担。没有退货的购买价格。

Answer 2

current.next.next = null;

应该是

current.next = null;

并且在尝试删除空List的第一个元素

时，NPE的实现失败

Answer 3

while(current.next.next != null)一直前进，直到它成真为止。此时您设置current.next.next = null;。它已经是。

Answer 4

到目前为止，Carl Smotricz的回答是最合乎逻辑的，当然其他人也做得很好解释。这是我的解决方案：

从当前节点开始并将其指向头
当两个节点当前节点的头部不为空时，则将当前节点设置为下一个节点
在结尾处将currtentnode的Next节点设置为null，换句话说，将最后一个节点设置为null /（将其删除）

代码

 Node currentNode = head; // start with current node and point it to head
    while (currentNode.next.next != null) // while the current next next  node is not null then set our  current node to next node
    {
        currentNode = currentNode.next;
    }
    // at the end set the Next node to the currtentnode to null
    // in other words, set the last node to null/ (remove it)
    currentNode.next = null;

这是视觉

currentNode     currentNode.next    currentNode.next.next
(head)                              (last Node)                     
    +-+-+         +-+-+                  +-+-+           
    | | +-------> | | +--------------->  | | |           
    +-+-+         +-+-+                  +-+-+

Answer 5

试试这个：

public class Node {

private int data; //holds an arbitrary integer
private Node next; //reference to the next node

public Node(int d,Node n)
{
    data=d;
    next=n;
}

// these methods let us use our variables
public void setNext(Node n)
{
    next=n;
}

public void setData(int d)
{
    data=d;
}
public Node getNext()
{
    return next;
}

public int getData()
{
    return data;
}
private static Node firstNode; //a reference to the first Node
private static Node lastNode=null; //a reference to the last Node

public static void display() //displays all the data in nodes
{
    Node n=firstNode;
    while(n!=null) // loops forward displaying 
    {
        System.out.print(n.getData()+",  ");
        n=n.getNext(); //move to the next node in the list
    }
}

public static void create(int d) //inserts the node into the list
{
    Node n =new Node(d, null); // create node
    if(lastNode!=null) // if it is not the first node
    {
        lastNode.setNext(n);
        lastNode=n;
    }
    else //if n is the first node
    {
        firstNode=n;
        lastNode=n;
    }
}


 public static void removeLast(){
        if (firstNode == null || firstNode.next == null){
            firstNode = null;
            return;
        }
        Node current = firstNode;
        Node tmphead = current;
        while(current.next.next != null){
            current = current.next;
        }
        current.next = null;
        firstNode = tmphead ;
    }


 public static void removeFirst(){
     if(firstNode!=null)
     {
        firstNode= firstNode.next;
     }
    }

public static void main(String[] args) {

    create(10);
    create(20);
    create(30);
    create(40);
    create(50);
    create(60);
    display();
    System.out.println();
    removeLast();
    display();
    System.out.println();
    removeFirst();
    display();
}
}

链表数据结构的java实现

5 个答案: