链表数据结构的java实现

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，但无法通过使用removeLast()方法删除最后一个元素。

我的链表节点类: LLNode类{ String value；LLNode next；

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

}

我的包含head节点的链表类：

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 

尽管如此，它还是必须这样：

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

我的实现出了什么问题？

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，你可能已经把自己弄得太复杂了。下面是我将如何使用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;
    }
  }
}

未经测试的！使用的风险自负。购买价格不退款。

Answer 2

current.next.next = null；

应该是

current.next = null；

并且当尝试删除空列表的第一个元素时，实现会失败，并返回NPE

Answer 3

while(current.next.next != null)不断进步，直到它成为真的。在这一点上设置current.next.next = null;。事实已经是这样了。