java– 根据另一个的排序对ArrayList进行排序

概述我在一个类(GlobalDataHolder.cardNameAndDescription)中有一个ArrayList<String>(),我想按字母顺序排序,但我还想在排序前面提到的arrayList时对ArrayList<Integer>()(UserBoxGlbImageAdapter.mGLBIcons)进行排序.旁注：两个数组的大小均为3(0-2).我正在编写自己的自定义compare

我在一个类(globalDataHolder.cardnameAndDescription)中有一个ArrayList< String>(),我想按字母顺序排序,但我还想在排序前面提到的arrayList时对ArrayList< Integer>()(UserBoxGlbImageAdapter.mGLBIcons)进行排序.

旁注：两个数组的大小均为3(0-2).

我正在编写自己的自定义compare()方法来执行此 *** 作,但我没有实现我正在寻找的东西.当我单击运行排序代码的按钮时,除非我单击按钮,否则无法实现正确的顺序3次虽然String ArrayList确实按字母排序.所以我认为我只需要对数组进行多次排序,因为数组的大小是(所以3次).

总而言之,String和Integer数据应该是相同的顺序,因为它们依赖于它的另一个但是我不能让它对两个数组都有效.

这些都没有效果.有人可以通过第二个阵列的排序告诉我这里我做错了什么吗？这是我的代码：

public class SortingDialog extends DialogFragment {@NonNull@OverrIDepublic Dialog onCreateDialog(Bundle savedInstanceState) {    // Create a builder to make the dialog building process easIEr    AlertDialog.Builder builder = new AlertDialog.Builder(getActivity());    builder.setTitle("Sorting Dialog");    builder.setSingleChoiceItems(R.array.sorting_options, 0,            new DialogInterface.OnClickListener() {                @OverrIDe                public voID onClick(DialogInterface dialogInterface, int i) {                    if (i == 1) {                        Toast.makeText(getActivity(), "2nd option Clicked", Toast.LENGTH_SHORT).show();                        if (getActivity().getSupportFragmentManager().findFragmentByTag("GLOBAL_FRAGMENT") != null) {                            sortGlobalListsBasedOnnameAndDesc();                        }                    }                    for (int j = 0; j < globalDataHolder.cardnameAndDescription.size(); j++) {                        Log.v("card_names", globalDataHolder.cardnameAndDescription.get(j));                    }                }            });    builder.setPositivebutton("OK", new DialogInterface.OnClickListener() {        @OverrIDe        public voID onClick(DialogInterface dialogInterface, int i) {            createtoast();            dismiss();        }    });    builder.setNegativebutton("Cancel", new DialogInterface.OnClickListener() {        @OverrIDe        public voID onClick(DialogInterface dialogInterface, int i) {            dismiss();        }    });    return builder.create();}private voID sortGlobalListsBasedOnnameAndDesc() {    if (Build.VERSION.SDK_INT >= Build.VERSION_CODES.N) {        globalDataHolder.cardnameAndDescription.sort(new Comparator<String>() {            @OverrIDe            public int compare(String s1, String s2) {                int ID1 = globalDataHolder.cardnameAndDescription.indexOf(s1);                int ID2 = globalDataHolder.cardnameAndDescription.indexOf(s2);                if (s1.equals(s2)) {                    return 0;                } else if (s1.comparetoIgnoreCase(s2) > 0) { //s1 is greater                    //Collections.swap(UserBoxGlbImageAdapter.mGLBIcons,ID2,ID1);                    swap(UserBoxGlbImageAdapter.mGLBIcons,ID2,ID1);                    swap(globalDataHolder.cardnameAndDescription,ID2,ID1);                    Log.d("case1","Called 1 time");                    return 1;                } else if (s1.comparetoIgnoreCase(s2) < 0) { //s1 is smaller                    //Collections.swap(UserBoxGlbImageAdapter.mGLBIcons,ID1,ID2);                    swap(UserBoxGlbImageAdapter.mGLBIcons,ID1,ID2);                    swap(globalDataHolder.cardnameAndDescription,ID1,ID2);                    Log.d("case2","Called 1 time");                    return -1;                } else {                    return 0;                }            }        });    }}private voID swap(List List,int objIndex1, int objIndex2) {    for (int i=0;i < List.size(); i++) {        Collections.swap(List,objIndex1,objIndex2);        UserBoxGlbImageAdapter.refreshFragmentVIEw(UserBoxGLBFragment.getUserBoxAdapter());    }}private voID createtoast() {    Toast.makeText(getActivity(), "Cards sorted based on AVG Stats", Toast.LENGTH_SHORT).show();}}

解决方法:

排序索引列表而不是列表本身并不困难.从那里你可以轻松地重新排序列表.

public class Test {    List<String> testStrings = Arrays.asList(new String[]{"One", "Two", "Three", "Four", "Five", "Six", "Seven", "Eight", "Nine", "Ten"});    List<Integer> testNumbers = Arrays.asList(new Integer[]{1, 2, 3, 4, 5, 6, 7, 8, 9, 10});    static <T extends Comparable<T>> List<Integer> getSortOrder(List<T> List) {        // Ints in increasing order from 0. One for each entry in the List.        List<Integer> order = IntStream.rangeClosed(0, List.size() - 1).Boxed().collect(Collectors.toList());        Collections.sort(order, new Comparator<Integer>() {            @OverrIDe            public int compare(Integer o1, Integer o2) {                // Comparing the contents of the List at the position of the integer.                return List.get(o1).compareto(List.get(o2));            }        });        return order;    }    static <T> List<T> reorder(List<T> List, List<Integer> order) {        return order.stream().map(i -> List.get(i)).collect(Collectors.toList());    }    public voID test() {        System.out.println("The strings: " + testStrings);        List<Integer> sortOrder = getSortOrder(testStrings);        System.out.println("The order they would be if they were sorted: " + sortOrder + " i.e. " + reorder(testStrings, sortOrder));        List<Integer> reordered = reorder(testNumbers, sortOrder);        System.out.println("Numbers in Alphabetical order of their names: " + reordered);    }    public static voID main(String[] args) {        new test().test();        System.out.println("Hello");    }}

打印：

The strings: [One, Two, Three, Four, Five, Six, Seven, Eight, Nine, Ten]

The order they would be if they were sorted: [7, 4, 3, 8, 0, 6, 5, 9, 2, 1] i.e. [Eight, Five, Four, Nine, One, Seven, Six, Ten, Three, Two]

Numbers in Alphabetical order of their names: [8, 5, 4, 9, 1, 7, 6, 10, 3, 2]

如果您觉得需要,可以随意添加自定义比较器.

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