使用λ:

    new AlertDialog.Builder(this).setMessage(getString(R.string.exit_msg))
        .setTitle(getString(R.string.info))
        .setPositiveButton(getString(R.string.yes), (arg0, arg1) -> {
            moveTaskToBack(true);
            finish();
        })
        .setNegativeButton(getString(R.string.no), (arg0, arg1) -> {
        })
        .show();

你还需要在gradle.build中设置支持java 8的级别语言:

compileOptions {
       targetCompatibility 1.8
       sourceCompatibility 1.8
}

使用λ:

    new AlertDialog.Builder(this).setMessage(getString(R.string.exit_msg))
        .setTitle(getString(R.string.info))
        .setPositiveButton(getString(R.string.yes), (arg0, arg1) -> {
            moveTaskToBack(true);
            finish();
        })
        .setNegativeButton(getString(R.string.no), (arg0, arg1) -> {
        })
        .show();

你还需要在gradle.build中设置支持java 8的级别语言:

compileOptions {
       targetCompatibility 1.8
       sourceCompatibility 1.8
}

我更喜欢双击后退按钮退出，而不是退出对话框。

在这个解决方案中，当第一次返回时，它会显示一个吐司，警告另一个回按将关闭应用程序。在这个例子中，不到4秒。

private Toast toast;
private long lastBackPressTime = 0;

@Override
public void onBackPressed() {
  if (this.lastBackPressTime < System.currentTimeMillis() - 4000) {
    toast = Toast.makeText(this, "Press back again to close this app", 4000);
    toast.show();
    this.lastBackPressTime = System.currentTimeMillis();
  } else {
    if (toast != null) {
    toast.cancel();
  }
  super.onBackPressed();
 }
}

令牌来自:http://www.androiduipatterns.com/2011/03/back-button-behavior.html

如果你不确定调用“back”是否会退出应用程序，或者将用户带到另一个活动，你可以将上述答案包装在一个检查中，isTaskRoot()。如果您的主活动可以多次添加到后台堆栈，或者如果您正在操作后台堆栈历史，就会发生这种情况。

if(isTaskRoot()) {
    AlertDialog.Builder builder = new AlertDialog.Builder(this);
    builder.setMessage("Are you sure you want to exit?")
       .setCancelable(false)
       .setPositiveButton("Yes", new DialogInterface.OnClickListener() {
           public void onClick(DialogInterface dialog, int id) {
                YourActivity.super.onBackPressed;
           }
       })
       .setNegativeButton("No", new DialogInterface.OnClickListener() {
           public void onClick(DialogInterface dialog, int id) {
                dialog.cancel();
           }
       });
    AlertDialog alert = builder.create();
    alert.show();

} else {
    super.onBackPressed();
}

我喜欢@GLee的方法，并像下面这样使用片段。

@Override
public void onBackPressed() {
    if(isTaskRoot()) {
        new ExitDialogFragment().show(getSupportFragmentManager(), null);
    } else {
        super.onBackPressed();
    }
}

使用片段的对话框:

public class ExitDialogFragment extends DialogFragment {

    @Override
    public Dialog onCreateDialog(Bundle savedInstanceState) {
        return new AlertDialog.Builder(getActivity())
            .setTitle(R.string.exit_question)
            .setPositiveButton(android.R.string.yes, new DialogInterface.OnClickListener() {
                @Override
                public void onClick(DialogInterface dialog, int which) {
                    getActivity().finish();
                }
            })
            .setNegativeButton(android.R.string.no, new DialogInterface.OnClickListener() {
                @Override
                public void onClick(DialogInterface dialog, int which) {
                    getDialog().cancel();
                }
            })
            .create();
    }
}

Just put this code in your first activity 

@Override
    public void onBackPressed() {
        if (drawerLayout.isDrawerOpen(GravityCompat.END)) {
            drawerLayout.closeDrawer(GravityCompat.END);
        }
        else {
// if your using fragment then you can do this way
            int fragments = getSupportFragmentManager().getBackStackEntryCount();
            if (fragments == 1) {
new AlertDialog.Builder(this)
           .setMessage("Are you sure you want to exit?")
           .setCancelable(false)
           .setPositiveButton("Yes", new DialogInterface.OnClickListener() {
               public void onClick(DialogInterface dialog, int id) {
                    finish();
               }
           })
           .setNegativeButton("No", null)
           .show();


            } else {
                if (getFragmentManager().getBackStackEntryCount() > 1) {
                    getFragmentManager().popBackStack();
                } else {

           super.onBackPressed();
                }
            }
        }
    }