linkedhashmap

If LinkedHashMap's time complexity is same as HashMap's complexity why do we need HashMap? What are all the extra overhead LinkedHashMap has when compared to HashMap in Java? LinkedHashMap will take more memory. Each entry in a normal HashMap just has the key and the value. Each LinkedHashMap entry has those references and references to the next and previous entries. There's also a little bit more housekeeping to do, although that's usually irrelevant. If LinkedHashMap's time complexity is same as HashMap's complexity why do we need HashMap? You should not confuse complexity with performance.

I was trying to implement a LRU cache using LinkedHashMap. In the documentation of LinkedHashMap ( http://docs.oracle.com/javase/7/docs/api/java/util/LinkedHashMap.html ), it says: Note that insertion order is not affected if a key is re-inserted into the map. But when I do the following puts public class LRUCache<K, V> extends LinkedHashMap<K, V> { private int size; public static void main(String[] args) { LRUCache<Integer, Integer> cache = LRUCache.newInstance(2); cache.put(1, 1); cache.put(2, 2); cache.put(1, 1); cache.put(3, 3); System.out.println(cache); } private LRUCache(int size) {

To sort it int ascending order I can use: myMap.entrySet().stream() .sorted(Map.Entry.comparingByValue()) .collect(Collectors.toMap(Entry::getKey, Entry::getValue)); How can I do it in decreasing order? Since Java 1.8 java.util.Comparator.reversed() myMap.entrySet().stream() .sorted(Map.Entry.comparingByValue().reversed()) .collect(Collectors.toMap(Entry::getKey, Entry::getValue)); To sort in reverse order, pass Comparator.reverseOrder() as parameter to comparingByValue . To get a LinkedHashMap , you must specifically request one with the 4-argument toMap() . If you don't specify what kind of

I'm Workin with Mongo using Jongo, when I do a query I receive a LinkedHashMap as result. Iterator one = (Iterator) friends.find(query).projection("{_id:0}").as(Object.class); while (one.hasNext()) { LinkedHashMap data= new LinkedHashMap(); data= (LinkedHashMap) one.next(); String content=data.toString(); } the problem is that if the json is {"user":"something"} content will be {user=something}, it is not a json is only toString method from HashMap. How I can get the original JSON?. I don't have a class to map the response, and it is not a solution create a map class, that is why I use a

I have following LinkedHashMap declaration. LinkedHashMap<String, ArrayList<String>> test1 my point is how can i iterate through this hash map. I want to do this following, for each key get the corresponding arraylist and print the values of the arraylist one by one against the key. I tried this but get only returns string, String key = iterator.next().toString(); ArrayList<String> value = (ArrayList<String> )test1.get(key) for (Map.Entry<String, ArrayList<String>> entry : test1.entrySet()) { String key = entry.getKey(); ArrayList<String> value = entry.getValue(); // now work with key and

I read that HashMap has the following implementation: main array ↓ [Entry] → Entry → Entry ← linked-list implementation [Entry] [Entry] → Entry [Entry] [null ] So, it has an array of Entry objects. Questions: I was wondering how can an index of this array store multiple Entry objects in case of same hashCode but different objects. How is this different from LinkedHashMap implementation? Its doubly linked list implementation of map but does it maintain an array like the above and how does it store pointers to the next and previous element? So, it has an array of Entry objects. Not exactly. It