As an option, we sort the collections and compare them:Collections.sort(list); Collections.sort(list001); System.out.println(list.equals(list001)); Help.

First of all, please note that in the line:myList.add((HashMap<String, String>) hmap); You put all the HashMap on the list, not one line. That may be what you need, but it's likely that you were planning on keeping one record here.Now look at the cycle you fill HashMap. On each internal cycle you place the current meaning of the Date and Name keys. And this is repeated on every external cycle iteration. I mean, first you put the name on the key, Moskow, then Kiev, and so on to Piter. And the key for each iteration changed to current and kept the final value when the cycles came out. And while you put HashMap in the List at every step, it's still the same reference as your log.UPDATE:If you're required to keep HashMap on the List, you're gonna need to re-establish it on every external cycle: for (int i = 0; i < jsonArray.length(); i++) { JSONObject objectInArray = jsonArray.getJSONObject(i); Iterator key = objectInArray.keys(); // !!! hmap = new HashMap&lt;String, String&gt;(); // !!! while (key.hasNext()) { String k = key.next().toString(); hmap.put(k,objectInArray.getString(k)); } myList.add((HashMap&lt;String, String&gt;) hmap); }

Tell me what's not clear.We're moving the body where the key is the last name, and the meaning is data on the user.Create a new $newlastnames.Let's start moving the cycle into a force.$firstname We're assigning a user name.Then let's get the data into our new body. $newlastnamesusing the name $firstname as a key, and as a matter of importance, we're putting data from the fields. firstname♪ age♪ sex♪

Works for masses of the same and different lengths:public static ArrayList function(ArrayList a, ArrayList b) { a.ensureCapacity(a.size() + b.size()); for (int i = 0; i < b.size(); i++) { int index = Math.min(i*2+1, a.size()); a.add(index, b.get(i)); } return a; } Scenario:public static ArrayList function2(ArrayList a, ArrayList b) { a.addAll(b); for (int i = b.size() - 1; i >= 0; i--) { int index = i*2 + 1; a.set(index, b.get(i)); a.set(index-1, a.get(i)); } return a; }

List<List<Object>> data - it's a two-strike. It's easier to imagine this as a routine table with lines and columns.So, data.get(0) - that's the first line. data.get(0).get(0) - The first line and the first column, that is [0;0] The cell is in the table.

ComparisonArrayList ArList = new ArrayList(); andCollection<String> collection = new ArrayList<String>(); We'll find small differences:new ArrayList() That's the same thing. new ArrayList<Object>()For ArList The type of reference will be https://docs.oracle.com/javase/7/docs/api/java/util/ArrayList.html A for collection The type of reference will be the interface. https://docs.oracle.com/javase/7/docs/api/java/util/Collection.html from which all collections have been established: lists, associations, many.You'll find out in more detail. http://docs.oracle.com/javase/7/docs/index.html ♪

When the second activate is launched, put your list in the interior as follows:intent.putStringArrayListExtra("key", arrayList); After, in the second activist, you get them from the intent so:List<String> list = getIntent().getStringArrayList("key");

You have a mistake in addPerson2 compared to:if ... && person.secondName.equals(book.get(i).lastName) && ... (sighs)secondName Comparison. lastNameYou don't need to add up on the list, you just have to check, and if you found a duplicate, fight and come out (mighty return). Adding is not necessary for every test within the cycle, but only when the entire list has been passed and it has been ascertained that there is no duplicate -- that is, after the cycle.The body of the method shall look such (short pattern): for (Person p: list) if (p.isTheSameThat(newPerson)) { System.out.println("Не катит!"); return; } list.add(newPerson); And, by the way, a match-check will be a good one.

That's what we should do.for(int _i = 0; _i < data.size(); _i++) { String msg = data.get(_i).getMsg(); String old_msg = data.get(_i).getMsg_id(); if (old_msg.equals("0")) { for (int i = 0; i &lt; newData.size(); i++) { remove(data.get(_i)); Log.d(LOG_TAG, "Удалено" + i + "-&gt;" + msg); } } }

Try this: intent.putStringArrayListExtra("items_to_parse", new ArrayList<>()); You may ask the size or fill it immediately, for example:intent.putStringArrayListExtra("items_to_parse", new ArrayList<>(Arrays.asList("Первый", "Второй", "Третий")));

Set up a method addData(List<ResponseMsgArray> dataToAdd)Use it by adding data to the list innerAdapterData.addAll(dataToAdd);As soon as this method is called, notify the adapter of its changes by method notifyDataSetChanged()

You need LinkedHashSet to keep unique values in the order in which you add them.List<String> list=new ArrayList<>("a","a","b"); Set<String> set=new LinkedHashSet<>(list); //внутри сета теперь ,"a","b"

How do you write the test initialized or not? and if not, initialize it.For example:public ArrayList<Part> initArrayList(ArrayList<Part> parts) { if (parts == null) { return new ArrayList<>(); } return parts; } UPD♪In view of the level of the question raised, he did not specifically want to confuse the author. http://sernam.ru/book_java.php?id=16 But if you know about him, the method. initArrayList(...) A few shorter can be achieved:public static ArrayList<Part> initArrayList(ArrayList<Part> parts) { return parts == null ? new ArrayList<>() : parts; } Actually, it is. syntactic sugarNot much.

Try to move the mass and take all 'target' for the chosen 'source':var Node = 1; var res = []; for(var index in edges) { if (edges[index].source == Node) res.push(edges[index].target); } var ret = {}; ret[Node] = res; If you need all the peaks with all the ribs, then:var res = {}; for(var node in nodes) { var rel = []; for(var edge in edges) { if (edges[edge].source == nodes[node].index) rel.push(edges[edge].target); } res[nodes[node].index] = rel; } A further option to obtain the masses of associated peaks for all peaks of the row (with one bypass):var res = {}; for(var edge in edges) { if (!res.hasOwnProperty(res[edges[edge].source].toString())) { res[edges[edge].source] = []; } res[edges[edge].source].push(edges[edge].target); }

Repeat values are shown because there is no default. When you create the element, try to set them empty values first and then process them with help. if...else

In order to keep a list of intervals, it may be:Use class https://developer.android.com/reference/android/util/Pair.html Conformity Mapwhere the keys are.key) shall be representative of the values (value) Example with Map:Map<String, Integer> urlColors = new HashMap<>(); urlColors.add( BASE + "CqmBjo5" + EXT, Color.parseColor("#ff0000"));

Apply to the row in the split sheet, thus breaking the line on the rig. Then you can work with this body or write another sheet.Example of use:String[] str = list.get(0).split(",");

Can you use Set? If objects are unique, Set s2 = ArrayList2.stream().collect(Collectors.toSet()); Set s1 = ArrayList1.stream().collect(Collectors.toSet()); s2.removeAll(s1); And what's left in s2 is added to ArrayList3

It's because https://libgdx.badlogicgames.com/nightlies/docs/api/com/badlogic/gdx/utils/Array.html#iterator-- returns the same object every time. And to irradiate the same body when you're already losing it (i.e. while in while ) you need to be used. https://libgdx.badlogicgames.com/nightlies/docs/api/com/badlogic/gdx/utils/Array.ArrayIterator.html as proposed by Array.iterator().It's done to keep you from creating a new object every time you've lost the body. I recommend that Array.iterator(s) be used for the internal cycle and ArrayIterator be used for the outside. Or keep the terator and "resue him every time.

Method get(i) facility nums type ArrayList<int[]> returns i- List element numswhich object is type int[] - a bunch of whole numbers. Furthermore, in the internal cycle this mass is initiated by values.