Java 使用“插入”和“删除”预定义的链接列表的“合并到”功能

好的，我对mergeTo函数有点问题，它应该把两个链表合并成一个（这个链表和那个链表-合并成这个）。以下是我的mergeTo代码：

public void mergeTo(WordLinkedList that){

   Node current = that.firstnode;
   int i = 0;

   while (current != null){ //iterating through all the elements of 'that' list
       this.insert(current.word);//inserting each successive word into 'this'
       that.remove(i); //removing each successive word from 'that'
       current = current.next;//moving the pointer forward
       i++; //moving the counter forward (for the remove function
   }
}

我对该函数的逻辑是迭代该链表的每个元素——将每个连续元素插入“this”列表，然后立即将其删除（使用i作为计数器）

这是我的插入函数：

public void insert(String newword){

  Node newNode = new Node(newword);
  size++;
  Node previous = null;
  Node present = firstnode;

  while ((present != null)&& (newword.compareTo(present.word) > 0)){
      previous = present;
      present = present.next;
  }
  if (present != null && newword.compareTo(present.word)==0){
      size--;
      return;
  }
  if ((previous == null)){
      firstnode = newNode;
  }
  else{
      previous.next = newNode;
  }
  newNode.next = present;
}

public String remove (int i){
   Node present = firstnode;
   Node previous = firstnode;

   if (i > this.size || i < 0)
       throw new IndexOutOfBoundsException("Hahahahahahaha it's out of bounds!!!!!!!!");

   if (i == 0 && firstnode != null){
       firstnode = firstnode.next;
   }
   else{
   for (int j = 0; j < i; j++){
       present = present.next;
   }
   for (int k = 0; k < i-1; k++){
       previous = previous.next;
   }

   previous.next = present.next;
   present.next = null;
   size--;
   }
   return present.word;
}

    String[] words6={"eta","gamma", "zeta"};//to test mergeTo
    String[] words7={"alpha","beta","phi"};//to test mergeTo - no common    words with words6
    listObj1=new WordLinkedList(words6);
    listObj2=new WordLinkedList(words7);
    listObj1.mergeTo(listObj2);
    System.out.println("this list: \n"+listObj1.toString());
    System.out.println("size of this list = " + listObj1.getSize());
    System.out.println("that list: \n"+listObj2.toString());
    System.out.println("size of that list = " + listObj2.getSize());

我的insert函数应该在保持字母顺序的同时将内容插入到我的链表中。我已经测试过了，我认为它可以正常工作。我不相信有什么问题

以下是我的删除功能：

public void insert(String newword){

  Node newNode = new Node(newword);
  size++;
  Node previous = null;
  Node present = firstnode;

  while ((present != null)&& (newword.compareTo(present.word) > 0)){
      previous = present;
      present = present.next;
  }
  if (present != null && newword.compareTo(present.word)==0){
      size--;
      return;
  }
  if ((previous == null)){
      firstnode = newNode;
  }
  else{
      previous.next = newNode;
  }
  newNode.next = present;
}

public String remove (int i){
   Node present = firstnode;
   Node previous = firstnode;

   if (i > this.size || i < 0)
       throw new IndexOutOfBoundsException("Hahahahahahaha it's out of bounds!!!!!!!!");

   if (i == 0 && firstnode != null){
       firstnode = firstnode.next;
   }
   else{
   for (int j = 0; j < i; j++){
       present = present.next;
   }
   for (int k = 0; k < i-1; k++){
       previous = previous.next;
   }

   previous.next = present.next;
   present.next = null;
   size--;
   }
   return present.word;
}

    String[] words6={"eta","gamma", "zeta"};//to test mergeTo
    String[] words7={"alpha","beta","phi"};//to test mergeTo - no common    words with words6
    listObj1=new WordLinkedList(words6);
    listObj2=new WordLinkedList(words7);
    listObj1.mergeTo(listObj2);
    System.out.println("this list: \n"+listObj1.toString());
    System.out.println("size of this list = " + listObj1.getSize());
    System.out.println("that list: \n"+listObj2.toString());
    System.out.println("size of that list = " + listObj2.getSize());

还有我的假设输出： 测试17-合并到 此列表： 阿尔法 贝塔 希腊字母表的第7个字母 伽马射线 phi 泽塔 此列表的大小=6 该清单： 列表是空的 该列表的大小=0

以及我的实际产出： 此列表： 阿尔法 贝塔 希腊字母表的第7个字母 伽马射线 泽塔 此列表的大小=5 该清单： 贝塔

---

该列表的大小=2

好的，因此我找到了解决问题的方法：）。在mergeTo方法中，应该将that.remove（i）更改为that.remove（0）-remove方法每次都删除第一个索引