就像赛义夫·阿西夫在回答中提到的那样，你可以使用另一个数据结构来跟踪id。
bitset是一种方法，另一种方法是使用treeset跟踪您分配和撤销的id，这将使id保持有序
例如

public class IdTracker {
    private TreeSet<Long> available;
    private TreeSet<Long> current;
    public IdTracker() {
        this.available = new TreeSet<Long>();
        this.current = new TreeSet<Long>();
    }
    public long getNextId() {
        //Check to see if this is the first time being called, setting initial id to 1
        if (available.isEmpty() && current.isEmpty()) {
            current.add(1L);
            return 1L;
        }
        //Check to see if we have any available values to use
        if (!available.isEmpty()) {
            //Remove from available and assign to current
            Long availableId = available.first();
            available.remove(availableId);
            current.add(availableId);
            return availableId;
        }
        //There are no available id's, get the highest current id and increment
        Long highestCurrentId = current.last();
        Long nextId = highestCurrentId + 1;
        current.add(nextId);
        return nextId;
    }
    public void removeId(long id) {
        //Remove from the current (if there) and place into available
        if (current.remove(id)) {
            available.add(id);
        } else {
            //Handle your failure case
        }
    }
}

所以在你的例子中

private List<Track> tracks;
private IdTracker idTracker;
public void addTrack(Track track) {
    long id = idTracker.getNextId();
    track.setId(id);
    this.tracks.add(track);
}
public void removeTrack(Track track) {
    if (tracks.contains(track)) {
        this.tracks.remove(track);
        this.idTracker.removeId(track.getId());
    } else {
        Terminal.printError("track with id " + track.getId() + " doesn't exist.");
        }
    }

一种方法是在另一个数据结构（可能是位集）中跟踪每个分配的id，特别是位集#nextclearbit（int）方法
所以每次你把东西放进 List<Tracks> ，则在位集中设置相对索引，并在 Track 已删除。
像下面这样

BitSet b = new BitSet();
// set the bits while adding tracks
b.set(0);
b.set(1);
b.set(2);
b.clear(1); // some track gets removed, so unset the bit
System.out.println(b); // {0, 2}
System.out.println(b.nextClearBit(0)); // 1