如何打印雙向鏈接列表的上一個節點和下一個節點？

Question

因此，當您播放專輯時，我正在創建CD播放器或iTunes跟蹤器的雙向鏈接列表。 使用AppendFront和AppendBack，我還使用了掃描儀和一條語句，該語句可以打印出曲目號，歌曲名稱以及上一首和下一首歌曲 。

我想我記下了代碼，但是，我對如何打印上一個節點和下一個節點感到困惑。 我想我的想法是將DoubleNode代碼與previous和next一起使用 ，但是如何使用它一直很困難。

//CD PROGRAM CODE
package dynamicData; //package is set

import java.util.Scanner; //imported the scanner to type

public class DoubleLinkedListTester {

public static void main(String[] args) {
    System.out.println("Welcome to your BeoCenter 2 CD Player.");
    System.out.println("Inserted CD: Beyoncé - I Am... Sasha Fierce (2008)");
    System.out.println("Tracklist: ");
    System.out.println("No. – Name");
    appendFront(1, "If I Were A Boy");
    appendBack(2, "Halo");
    appendBack(3, "Disappear");
    appendBack(4, "Broken-Hearted Girl");
    appendBack(5, "Ave Maria");
    appendBack(6, "Satellites");
    appendBack(7, "Single Ladies (Put a Ring on It)");
    appendBack(8, "Radio");
    appendBack(9, "Diva");
    appendBack(10, "Sweet Dreams");
    appendBack(11, "Video Phone [featuring Lady Gaga]");
    appendBack(12, "Ego [featuring Kanye West]");
    appendBack(13, "Roc");

    System.out.println(" ");
    //tracklist();
    Scanner album = new Scanner(System.in); //scanner is set
    System.out.print("Which track number would you like to hear?: "); //asking the use what their favourite song is
    int input = album.nextInt(); //would type in their input

    switch (input) {
        case 1 : scan(input, input, "If I Were A Boy");
            break;
        case 2 :  scan(input, input, "Halo");
            break;
        case 3 :  scan(input, input, "Disappear");
            break;
        case 4 :  scan(input, input, "Broken-Hearted Girl");
            break;
        case 5 :  scan(input, input, "Ave Maria");
            break;
        case 6 :  scan(input, input, "Satellites");
            break;
        case 7 :  scan(input, input, "Single Ladies (Put A Ring On It)");
            break;
        case 8 :  scan(input, input, "Radio");
            break;
        case 9 :  scan(input, input, "Diva");
            break;
        case 10 :  scan(input, input, "Sweet Dreams");
            break;
        case 11 :  scan(input, input, "Video Pohone [featuring Lady Gaga]");
            break;
        case 12 :  scan(input, input, "Ego [featuring Kanye West]");
            break;
        case 13 :  scan(input, input, "Roc");
            break;
            default : System.out.println("Invalid selection. Please restart program.");
            break;
        }

} //main void ends

private static DoubleNode headNode =  null;
private static DoubleNode tailNode =  null;
private static int size;

public int size() {
    return size;
    }

//add new node at the front of the DLL
public static void appendFront(int trackNumber, String songName) {
    DoubleNode newNode = new DoubleNode(trackNumber, songName, headNode, null);

    if(headNode != null ) {
        headNode.previous = newNode;
        } //end of if 1

    headNode = newNode;
    if(tailNode == null) {
        tailNode = newNode;
        } //end of if 2
    size++;
    System.out.println(trackNumber+ "   –   " + songName);
}

//add at the back
public static void appendBack(int trackNumber, String songName) {
    DoubleNode newNode = new DoubleNode(trackNumber, songName, headNode, null);

    if(tailNode != null ) {
        tailNode.previous = newNode;
        } //end of if 1

    tailNode = newNode;
    if(headNode == null) {
        headNode = newNode;
        } //end of if 2
    size++; //size is then added on
    System.out.println(trackNumber+ "   –   " + songName);

}

public static void tracklist() {
    DoubleNode currentNode = headNode;

    while (currentNode != null ) {

        System.out.println(currentNode.data + " — " + currentNode.description);
        System.out.println("Next Link: " + currentNode.next);
        currentNode = currentNode.next;
        System.out.println(" ");
    }
}

public static void scan(int input, int trackNumber, String songName) {

if (input >= 1 ) {
    System.out.println(" ");
    System.out.println("Song selected.");
    System.out.println("Now playing: ");
    System.out.println("Track " + trackNumber + " – " + songName);
    System.out.println("Previous song: " + songName);
    System.out.println("Next song: " + songName);
    System.out.println(" ");
    } else {
    System.out.println("Goodbye!");
} //end of if
}

//scan backwards
public static void scanBackward() {
    System.out.println("Scanning backwards through playlist.");

    DoubleNode temp = tailNode;

    while(temp != null){
        System.out.println(temp.data + " " + temp.description);
        temp = temp.previous;
    }
  }
} //class ends

下面是我正在使用的Node代碼。

//Node Code
package dynamicData; //package is set

public class DoubleNode { //class is set

public int data; //created a public integer called data to store the int number
public DoubleNode next; //created a public Node called next to call the next element or node in the array
public DoubleNode previous;
public String description; //created a public String called description to store the description of what the element or node is


public DoubleNode (int trackNumber, String songName, DoubleNode next, DoubleNode previous) { //the object is set as we recall the node class
    data = trackNumber; //we set the data as the track number we are on
    description = songName; //we set the description as the songs name to know what we are on
    this.next = next; //using this.next to make sure the computer is getting from this public node and not mixing it up or getting confused through the other next up above
    this.previous = previous;
    //this next is using the next element or node in the array
} //end of inner Node

public String toString(){ //public string to return the name of the site
    return description; //would return the description, which is the site's name
} //end of string
} //end of class Node

注意評論，我知道現在到處都是，我通常在完成代碼后就將其修復。 我感到困惑的是，是否仍然需要訪問上一個或下一個的DoubleNode代碼，還是仍然需要創建一個對象？

Answer 1

您似乎對要求您做什么感到困惑。 似乎這也可能是學校的作業，我們通常不為這里的人做家庭作業...

話雖如此，這是一個提示：

您的switch語句完全沒有必要並且是錯誤的。 您有一個歌曲列表，用戶正在輸入要播放的歌曲； 您應該掃描 （即遍歷）列表以搜索歌曲。 您的switch語句有效地復制了列表中已有的內容。

如果列表中有成千上萬首歌曲怎么辦？ 您會像以前一樣為每首歌寫一個case子句嗎？ 如果您事先不知道列表中的內容（可能是從數據庫中加載的）怎么辦？ 那么，您將如何編寫switch語句？

您根本不需要該開關。

您應該在列表中循環搜索所需的歌曲，即從列表的開頭開始，並在遍歷歌曲時對它們進行計數。

到達請求的歌曲后，它的節點同時具有上一個和下一個鏈接，因此在打印歌曲的標題時，您還可以查看相鄰節點並打印出它們的標題。

希望它能幫助您擺脫困境。

Answer 2

該代碼有很多錯誤。

將headNode和tailNode初始化為兩個虛擬節點要容易得多。

然后在您的appendFront ，應該執行此操作

headNode.previous = newNode;
newNode.next = headNode;

headNode = newNode;
headNode.previous = tailNode;
tailNode.next = headNode;

同樣，在您的appendBack

tailNode.next = newNode;
newNode.previous = tailNode;

tailNode = newNode;
tailNode.next = headNode;
headNode.previous = tailNode;

由於這是一個跟蹤列表，因此應該是一個循環：

headNode <-> node1 <-> node2 ... <-> headNode