我刚刚得到了解决方案，想把它贴在这里。这样可以帮助其他人
答案很简单，我得到了videoFile（TAG1）和link（TAG2），绕过了这种格式的链接

https://api.vimeo.com/me/videos/123456789

最后的代码是这样的

final String uri = "https://api.vimeo.com/me/videos/123456789";

而不是这个

final String uri = "https://player.vimeo.com/videos/123456789/config";

这里是我的完整代码，帮助我通过使用Vimeo网络库在Android应用程序中播放视频

展示最终代码

public class PlayActivity extends AppCompatActivity {

VideoView videoView;

@Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_play);

    videoView = findViewById(R.id.player);
    // Getting access Token

    String accessToken = getString(R.string.access_token);
    Configuration.Builder configBuilder = new Configuration.Builder(accessToken)
            .enableCertPinning(false);
    //Vimeo Client autenticated

    VimeoClient.initialize(configBuilder.build());
    // the video uri; if you have a video, this is video.uri

    final String uri = "https://api.vimeo.com/me/videos/123456789"; 

    VimeoClient.getInstance().fetchNetworkContent(uri, new ModelCallback<Video>(Video.class) {
    
    @Override
    public void success(Video video) {
        Toast.makeText(PlayActivity.this, "Sucessful" + video, Toast.LENGTH_SHORT).show();
       
        ArrayList<VideoFile> videoFiles = video.files;
        Log.i("TAG1", "videoFiles " + videoFiles);
        if (videoFiles != null && !videoFiles.isEmpty()) {
            VideoFile videoFile = videoFiles.get(0); // you could sort these files by size, fps, width/height
            
            String link = videoFile.getLink();
            Log.i("TAG2", "link " + link);
            // load link
            // use the link to play the video by **EXO Player** or **Video View**
           // Start your video player here
           }
         }

    @Override
    public void failure(VimeoError error) {

        Log.i("TAG3", "vimeo error  : " + error.getErrorMessage());
        Toast.makeText(PlayActivity.this, "failure due to " + error.getErrorMessage(), Toast.LENGTH_SHORT).show();

       }
    });
    }
}

我在库版本3. 12. 0上试过这个&它工作正常：-
请注意，这个访问令牌需要有所有三个访问被检查的时候，它的建设-“公共”，“私人”，“视频文件”

val confBuilder = VimeoApiConfiguration.Builder(accessToken)
val configuration = confBuilder.build()
Authenticator.initialize(configuration)
VimeoApiClient.initialize(configuration,Authenticator.instance())
VimeoApiClient.instance()
   .fetchVideo(finalUrl, null, null, null, object : VimeoCallback<Video>{
 //here url should be like "https://api.vimeo.com/videos/{video_id}" otherwise it wasn't working whatever the url was
     override fun onError(error: VimeoResponse.Error) {
        Logger.d("TAG", "vimeo error = ${error.message}")
        Toast.makeText(applicationContext, error.message, Toast.LENGTH_SHORT).show()
        }
    override fun onSuccess(response: VimeoResponse.Success<Video>) {
      val video = response.data
      Log.d("TAG", "vimeo video user = ${response.data.user?.name}")
        val play: Play? = video.play
        val progressiveFiles: List<ProgressiveVideoFile>? = play.progressive
      val linkToMp4File: String? = progressiveFiles?.get(0)?.link
                                runExoplayer(linkToMp4File)
       }
      })

在此播放中，如果视频是由您正在使用的访问令牌的同一用户创建的，则数据将存在。
您可以通过比较您正在进入视频对象的用户（上面记录的）和您已经验证的用户（如下所示）来验证这一点：-

VimeoApiClient.instance().fetchCurrentUser(null, null, object : VimeoCallback<User>{
        override fun onError(error: VimeoResponse.Error) {

        }
        override fun onSuccess(response: VimeoResponse.Success<User>) {
            Logger.d("TAG", "vimeo user = ${response.data.name}")
        }
    })

只有那些视频将播放数据的用户是相同的视频创建者/成员vimeo帐户作为您的访问令牌所有者。这是在官方文件中提到这里https://developer.vimeo.com/api/authentication（了解认证过程）