Java 使用带有LinkedList的compareTo方法按字母顺序添加对象

我已经创建了自己的LinkedList，其中包含一个歌曲对象（标题、艺术家等）。我试图添加一个功能，用户输入歌曲详细信息，并按艺术家的字母顺序将歌曲添加到列表中。我的问题是，我不知道如何设置if-else语句以按艺术家的字母顺序将其添加到列表中，提前感谢

歌曲类中的compareTo（）

//Should return negative if in order, positive if out of order, 0 if equal
public int compareTo(Song s) {
        return title.compareTo(s.getTitle()); //I think this is correct?
}

public static void addSong(LinkedList list){
    Scanner input = new Scanner(System.in);

    System.out.println("Enter song title: ");
    String title = input.nextLine();
    System.out.println("Enter song artist: ");
    String artist = input.nextLine();
    System.out.println("Enter song album: ");
    String album = input.nextLine();
    System.out.println("Enter song length: ");
    String length = input.nextLine();

    Song songInfo = new Song(title, artist, album, length);
    for (int i = 0; i < list.size(); i ++){
        if ( //compareTo should be used here < 0 ){
        }else if( //compareTo should be used here == 0)

    }

public class Node {
private Song song;
private Node next;
public Node( Song s ){
    song = s;
    next = null;
}
public Node( Song s, Node n ){
    song = s;
    next = n;
}
public Node getNext(){
    return next;
}
public void setNext(Node n){
    next = n;
}
public Song getValue(){
    return song;
}
public void setValue( Song s ){
    song = s;
}

主类

//Should return negative if in order, positive if out of order, 0 if equal
public int compareTo(Song s) {
        return title.compareTo(s.getTitle()); //I think this is correct?
}

public static void addSong(LinkedList list){
    Scanner input = new Scanner(System.in);

    System.out.println("Enter song title: ");
    String title = input.nextLine();
    System.out.println("Enter song artist: ");
    String artist = input.nextLine();
    System.out.println("Enter song album: ");
    String album = input.nextLine();
    System.out.println("Enter song length: ");
    String length = input.nextLine();

    Song songInfo = new Song(title, artist, album, length);
    for (int i = 0; i < list.size(); i ++){
        if ( //compareTo should be used here < 0 ){
        }else if( //compareTo should be used here == 0)

    }

public class Node {
private Song song;
private Node next;
public Node( Song s ){
    song = s;
    next = null;
}
public Node( Song s, Node n ){
    song = s;
    next = n;
}
public Node getNext(){
    return next;
}
public void setNext(Node n){
    next = n;
}
public Song getValue(){
    return song;
}
public void setValue( Song s ){
    song = s;
}

节点类

//Should return negative if in order, positive if out of order, 0 if equal
public int compareTo(Song s) {
        return title.compareTo(s.getTitle()); //I think this is correct?
}

public static void addSong(LinkedList list){
    Scanner input = new Scanner(System.in);

    System.out.println("Enter song title: ");
    String title = input.nextLine();
    System.out.println("Enter song artist: ");
    String artist = input.nextLine();
    System.out.println("Enter song album: ");
    String album = input.nextLine();
    System.out.println("Enter song length: ");
    String length = input.nextLine();

    Song songInfo = new Song(title, artist, album, length);
    for (int i = 0; i < list.size(); i ++){
        if ( //compareTo should be used here < 0 ){
        }else if( //compareTo should be used here == 0)

    }

public class Node {
private Song song;
private Node next;
public Node( Song s ){
    song = s;
    next = null;
}
public Node( Song s, Node n ){
    song = s;
    next = n;
}
public Node getNext(){
    return next;
}
public void setNext(Node n){
    next = n;
}
public Song getValue(){
    return song;
}
public void setValue( Song s ){
    song = s;
}

---

只需浏览列表，直到您发现第一个元素比您的元素大。然后在该元素之前添加元素。因此，您需要<代码>实现可重复的< /代码>，然后您需要实现<代码> Iterator <代码>，它支持<代码>添加< /代码>，在您的<代码>列表中。修改您的代码> Addio< /Cult>逻辑，以找到适当的索引，在这里插入“代码>节点< /代码>，而不是总是最后添加。好吧，我还是有点困惑。我不确定if语句是看起来像这样，还是完全关闭了if（artist.compareTo（list.get（I）.getValue（）.getArtist（））<0）**您永远不应该从

链接列表中
get（I）
；提示：在

List`上循环，并为每个元素调用get（i）——时间复杂度是多少？因此，您需要

实现Iterable

，并查看它将把您带到哪里。