如何從鏈表中刪除隨機選擇的元素？

Question

我正在嘗試對一個類進行分配，在該類中，我使用String Bag類的remove方法一次返回一個鏈表的所有元素，然后從列表中刪除該元素。 我有一個開始，但我不知道該怎么做。 有人可以幫忙嗎？

 public String remove()
  {
      Random rand = new Random();
      int randNum = rand.nextInt(numItems);
      //generate random number
      int count = 0;
      String get;
      currNode = firstNode;
      //temporary node to get String from

      while(count < randNum)
      {
          currNode = currNode.getLink();  
          count++;
      }
      //randomly select node to get String from
      get = currNode.getInfo();

      numItems--;
      if(numItems == 0)
      {
          firstNode = null;
      }
      //decrement the number of items in the bag and make the first node
      //null when it reaches 0
      return get;

  }

編輯：這是應用程序級別：

public class StringBagTest 
{

 public static void main(String[] args) 
 {          
    LLStringBag bag = new LLStringBag();
    bag.insert("Hat");
    bag.insert("Shirt");
    bag.insert("Pants");
    bag.insert("Shoes");
    //insert 4 strings into the list
    while(!bag.isEmpty())
    {
    System.out.println(bag.remove());
    }
    //randomly removes all contents of list
  }
}

Answer 1

如果要按索引刪除隨機選擇的元素，則它看起來像這樣：

public void removeRandomElement() {
        int index = new Random().nextInt(size);
        Node current = head;
        Node prev = head;
        for (int i = 0; i < index; i++) {
            prev = current;
            current = current.next;
        }
        prev.next = current.next;
        current.next = null;
        size--;
    }

對於單鏈列表，其中size是列表的當前大小，為head -head節點。

換句話說，您正在對所選元素執行以下操作：

Answer 2

如果要刪除鏈表的所有元素，則可以使用內置的clear（）方法。

如果您不想使用該方法，則可以將頭節點設置為null。 垃圾收集器將負責其余的工作。

如果您想要一次一次刪除一件東西的remove方法，而不關心它刪除了什么，我建議您只刪除找到的第一個元素。 如果它在鏈接列表中，則只需將臨時節點分配給頭節點，將頭節點重新分配給下一個節點，然后返回臨時節點。

Answer 3

看一下這個鏈接： link

也是一個完整的例子：（ 創建自己的鏈接和列表 ）

（下面的示例是一個具有（鏈接）的鏈接列表，它的鏈接是一個點，例如A（50,3）。您可以將其轉換為所需的任何內容...）

鏈接

public class DoublePoint {

public double X;
public double Y;
public int LinkKey=0;
public DoublePoint nextLink; //keeps the nextLink

//Constructor
public DoublePoint(double Xpos,double Ypos,int key){
    X=Xpos;
    Y=Ypos;
    LinkKey=key;
}



public void printLinkKey(){
    System.out.println(LinkKey);
}


 //Return Link key

public String returnLinkKey(){

    return ""+LinkKey;
}



 public void changeContent(double x,double y){
      X=x;
    Y=y;

  }

public void ChangeLinkKey(int key){
    LinkKey=key;

}

  }

名單：

public class ListDoublePoints {

 public DoublePoint first;
 public int key; 
 public int totalLinks=0; 

public ListDoublePoints(){
    first=null;
    key=0;
}


//Insert


public void insertLink(double x,double y){
   DoublePoint newLink = new DoublePoint(x,y,key);
   newLink.nextLink=first;
   first=newLink;
   key++;
   totalLinks++;
}



//Find


public DoublePoint findLinkAt(int key){
    DoublePoint current=first;

 while(current.LinkKey!=key){ 
     if(current.nextLink==null)
         return null;
     else
        current=current.nextLink;

 }
 return current;

}


//Delete using Link key (similar with remove(int position) with ready java lists)


public String deleteLinkAt(int linkKey){

    DoublePoint current =first;
    DoublePoint previous=first;

    while(current.LinkKey !=linkKey){
        if(current.nextLink == null ){
            return "boom";}
        else
            previous=current;
            current=current.nextLink;
    }

    if(current==first)
        first=first.nextLink;
    else
        previous.nextLink=current.nextLink;

    --totalLinks;

    return "ok";
}


//Return


public int  LinksNumber(){
     return totalLinks;
}



//Print


public void displayList(){
   DoublePoint current=first;
  while(current!=null){
     current.displayLink();
     current=current.nextLink;
   }

}


 public void displayTheNumberOfLinks(){
    System.out.println(totalLinks);
  }

}

*讓我知道您是否想要以上類似內容或

只是為了與Java准備就緒列表一起工作.. *

Answer 4

你的意思是這樣的嗎？？？

碼

 private LinkedList<String> list = new LinkedList<>();

    private void fillList() {
        for (int i = 0; i < 10; i++) {
            list.add("Hello " + i);
        }
    }

    private void removeAllRandomly() {

        Random random = new Random();

        while (!list.isEmpty()) {
            int randomPosition = random.nextInt(list.size());
            String s = list.remove(randomPosition);
            System.out.println(String.format("Item on position: %s (%s) was removed", randomPosition, s));
        }

    }

---

結果

Item on position: 9 (Hello 9) was removed
Item on position: 1 (Hello 1) was removed
Item on position: 1 (Hello 2) was removed
Item on position: 2 (Hello 4) was removed
Item on position: 5 (Hello 8) was removed
Item on position: 0 (Hello 0) was removed
Item on position: 3 (Hello 7) was removed
Item on position: 1 (Hello 5) was removed
Item on position: 1 (Hello 6) was removed
Item on position: 0 (Hello 3) was removed