我采用了@(Kirk Woll)建议的解决方案，它对我很有效。我还在“联系人”应用程序的Android源代码中看到，他们使用了类似的技术。我还想补充一些具体情况: 在我的listactivity派生类的顶部:

private static final String LIST_STATE = "listState";
private Parcelable mListState = null;

然后，一些方法重写:

@Override
protected void onRestoreInstanceState(Bundle state) {
    super.onRestoreInstanceState(state);
    mListState = state.getParcelable(LIST_STATE);
}

@Override
protected void onResume() {
    super.onResume();
    loadData();
    if (mListState != null)
        getListView().onRestoreInstanceState(mListState);
    mListState = null;
}

@Override
protected void onSaveInstanceState(Bundle state) {
    super.onSaveInstanceState(state);
    mListState = getListView().onSaveInstanceState();
    state.putParcelable(LIST_STATE, mListState);
}

当然，“loadData”是我从DB中检索数据并将其放入列表的函数。

在我的Froyo设备上，当你改变手机方向时，当你编辑一个项目并返回列表时，这都是有效的。

试试这个:

// save index and top position
int index = mList.getFirstVisiblePosition();
View v = mList.getChildAt(0);
int top = (v == null) ? 0 : (v.getTop() - mList.getPaddingTop());

// ...

// restore index and position
mList.setSelectionFromTop(index, top);

Explanation: ListView.getFirstVisiblePosition() returns the top visible list item. But this item may be partially scrolled out of view, and if you want to restore the exact scroll position of the list you need to get this offset. So ListView.getChildAt(0) returns the View for the top list item, and then View.getTop() - mList.getPaddingTop() returns its relative offset from the top of the ListView. Then, to restore the ListView's scroll position, we call ListView.setSelectionFromTop() with the index of the item we want and an offset to position its top edge from the top of the ListView.

If you are saving/restoring scroll position of ListView yourself you are essentially duplicating the functionality already implemented in android framework. The ListView restores fine scroll position just well on its own except one caveat: as @aaronvargas mentioned there is a bug in AbsListView that won't let to restore fine scroll position for the first list item. Nevertheless the best way to restore scroll position is not to restore it. Android framework will do it better for you. Just make sure you have met the following conditions:

确保你没有调用setSaveEnabled(false)方法，也没有为xml布局文件中的列表设置android:saveEnabled="false"属性 为ExpandableListView重写long getCombinedChildId(long groupId, long childId)方法，使其返回正长数(BaseExpandableListAdapter类中的默认实现返回负数)。下面是一些例子:

.

@Override
public long getChildId(int groupPosition, int childPosition) {
    return 0L | groupPosition << 12 | childPosition;
}

@Override
public long getCombinedChildId(long groupId, long childId) {
    return groupId << 32 | childId << 1 | 1;
}

@Override
public long getGroupId(int groupPosition) {
    return groupPosition;
}

@Override
public long getCombinedGroupId(long groupId) {
    return (groupId & 0x7FFFFFFF) << 32;
}

如果在一个片段中使用了ListView或ExpandableListView，不要在活动重新创建片段(例如在屏幕旋转后)。使用findFragmentByTag(String标签)方法获取片段。 确保ListView有一个唯一的android:id。

To avoid aforementioned caveat with first list item you can craft your adapter the way it returns special dummy zero pixels height view for the ListView at position 0. Here is the simple example project shows ListView and ExpandableListView restore their fine scroll positions whereas their scroll positions are not explicitly saved/restored. Fine scroll position is restored perfectly even for the complex scenarios with temporary switching to some other application, double screen rotation and switching back to the test application. Please note, if you are explicitly exiting the application (by pressing the Back button) the scroll position won't be saved (as well as all other Views won't save their state). https://github.com/voromto/RestoreScrollPosition/releases

最好的解决方案是:

// save index and top position
int index = mList.getFirstVisiblePosition();
View v = mList.getChildAt(0);
int top = (v == null) ? 0 : (v.getTop() - mList.getPaddingTop());

// ...

// restore index and position
mList.post(new Runnable() {
    @Override
    public void run() {
      mList.setSelectionFromTop(index, top);
   }
});

你必须在邮件和线程中调用!

如果你在一个活动上使用片段，你可以这样做:

public abstract class BaseFragment extends Fragment {
     private boolean mSaveView = false;
     private SoftReference<View> mViewReference;

     @Override
     public View onCreateView(LayoutInflater inflater, ViewGroup container, Bundle savedInstanceState) {
          if (mSaveView) {
               if (mViewReference != null) {
                    final View savedView = mViewReference.get();
                    if (savedView != null) {
                         if (savedView.getParent() != null) {
                              ((ViewGroup) savedView.getParent()).removeView(savedView);
                              return savedView;
                         }
                    }
               }
          }

          final View view = inflater.inflate(getFragmentResource(), container, false);
          mViewReference = new SoftReference<View>(view);
          return view;
     }

     protected void setSaveView(boolean value) {
           mSaveView = value;
     }
}

public class MyFragment extends BaseFragment {
     @Override
     public View onCreateView(LayoutInflater inflater, ViewGroup container, Bundle savedInstanceState) {
          setSaveView(true);
          final View view = super.onCreateView(inflater, container, savedInstanceState);
          ListView placesList = (ListView) view.findViewById(R.id.places_list);
          if (placesList.getAdapter() == null) {
               placesList.setAdapter(createAdapter());
          }
     }
}

我发布这篇文章是因为我很惊讶没有人提到这一点。

当用户单击返回按钮后，他将返回到列表视图，在相同的状态，因为他离开它。

这段代码将覆盖“向上”按钮的行为与后退按钮相同，所以在Listview ->细节->回到Listview(没有其他选项)的情况下，这是最简单的代码来维护滚动位置和Listview中的内容。

 public boolean onOptionsItemSelected(MenuItem item) {
     switch (item.getItemId()) {
         case android.R.id.home:
             onBackPressed();
             return(true);
     }
     return(super.onOptionsItemSelected(item)); }

注意:如果你可以从细节活动转到另一个活动，向上按钮将返回到该活动，所以你必须操作后退按钮历史，以使其工作。