Develop Reference

Is Set sorted in some manner by default?

I have this set with elements added in the given order.
Set<String> nations = new HashSet<String>();
nations.add("Australia");
nations.add("Japan");
nations.add("Taiwan");
nations.add("Cyprus");
nations.add("Cuba");
nations.add("India");
When I print the record -
for (String s : nations) {
System.out.print(s + " ");
}
It always gives this output in the order
Cuba Cyprus Japan Taiwan Australia India
As far as I know a Set is not sorted by default, but why do I get the same result in a particular sorted manner?
Update : Here is the actual question -
public static Function<String,String> swap = s -> {
if(s.equals("Australia"))
return "New Zealand";
else
return s;
};
Set<String> islandNations = Set.of("Australia", "Japan", "Taiwan", "Cyprus", "Cuba");
islandNations = islandNations.stream()
.map(swap)
.map(n -> n.substring(0, 1))
.collect(Collectors.toSet());
for(String s : islandNations){
System.out.print(s);
}
and answers one of these
CTJN
TJNC
TCNJ

HashSet's documentation says:
It makes no guarantees as to the iteration order of the set.
No guarantees means no guarantees. For example, it could be sorted order, reverse sorted order, random order, or sorted order except on Tuesdays when it's random.
(In practice, the iteration order is usually always the same for the same Java version, or at least for the same run of the JVM, and that order is produced by a deliberately convoluted algorithm based on the hash codes of the elements. However, if you depend on that behavior, it will usually change at the worst possible time.)

HashSet does not preserve the order of insertion of elements, as the order is maintained based on the hashing mechanism like Map because the add() method internally inserts the element as a key in a Map.
HashSet
//add method implementation for HashSet
public boolean add(E e) {
return map.put(e, PRESENT)==null;
}
here, map is a private transient HashMap<E,Object> map;
As the map is HashMap here, so no sorting will be done.
TreeSet
//add method implementation for TreeSet
public boolean add(E e) {
return m.put(e, PRESENT)==null;
}
here, map is a private transient NavigableMap<E,Object> m;
As the map is NavigableMap here, so sorting will be done.

The HashSet uses the hash value of each element for storage.
The important points about Java HashSet class are:
HashSet stores the elements by using a mechanism called hashing.
HashSet contains unique elements only.
HashSet allows null value.
HashSet class is non-synchronized.
HashSet doesn't maintain the insertion order. Here, elements are inserted on the basis of their hashcode.
HashSet is the best approach for search operations.
The initial default capacity of HashSet is 16, and the load factor is 0.75.
if you need to store (and display) in the order you can use the SortedSet interface, like this:
SortedSet<String> orderedList = new TreeSet<String>();
orderedList.add("C");
orderedList.add("D");
orderedList.add("E");
orderedList.add("A");
orderedList.add("B");
orderedList.add("Z");
for (String value : orderedList)
System.out.print(value + ", ");
Output:
A, B, C, D, E, Z,
Remembering: SortedSet uses the Comparable interface and the compareTo() method to sort the String values. If you have a customized class you should implement this interface/method to use in this approach.
Or, you can define the comparator that must be used:
SortedSet<Person> persons = new TreeSet<>Comparator.comparing(Person::getName));

How to sort a HashSet?

For lists, we use the Collections.sort(List) method. What if we want to sort a HashSet?

A HashSet does not guarantee any order of its elements. If you need this guarantee, consider using a TreeSet to hold your elements.
However if you just need your elements sorted for this one occurrence, then just temporarily create a List and sort that:
Set<?> yourHashSet = new HashSet<>();
...
List<?> sortedList = new ArrayList<>(yourHashSet);
Collections.sort(sortedList);

Add all your objects to the TreeSet, you will get a sorted Set. Below is a raw example.
HashSet myHashSet = new HashSet();
myHashSet.add(1);
myHashSet.add(23);
myHashSet.add(45);
myHashSet.add(12);
TreeSet myTreeSet = new TreeSet();
myTreeSet.addAll(myHashSet);
System.out.println(myTreeSet); // Prints [1, 12, 23, 45]
Update
You can also use TreeSet's constructor that takes a HashSet as a parameter.
HashSet myHashSet = new HashSet();
myHashSet.add(1);
myHashSet.add(23);
myHashSet.add(45);
myHashSet.add(12);
TreeSet myTreeSet = new TreeSet(myHashSet);
System.out.println(myTreeSet); // Prints [1, 12, 23, 45]
Thanks #mounika for the update.

Java 8 way to sort it would be:
fooHashSet.stream()
.sorted(Comparator.comparing(Foo::getSize)) //comparator - how you want to sort it
.collect(Collectors.toList()); //collector - what you want to collect it to
*Foo::getSize it's an example how to sort the HashSet of YourItem's naturally by size.
*Collectors.toList() is going to collect the result of sorting into a List the you will need to capture it with List<Foo> sortedListOfFoo =

You can use a TreeSet instead.

Use java.util.TreeSet as the actual object. When you iterate over this collection, the values come back in a well-defined order.
If you use java.util.HashSet then the order depends on an internal hash function which is almost certainly not lexicographic (based on content).

Just in-case you don't wanna use a TreeSet you could try this using java stream for concise code.
set = set.stream().sorted().collect(Collectors.toCollection(LinkedHashSet::new));

You can use Java 8 collectors and TreeSet
list.stream().collect(Collectors.toCollection(TreeSet::new))

Based on the answer given by #LazerBanana i will put my own example of a Set sorted by the Id of the Object:
Set<Clazz> yourSet = [...];
yourSet.stream().sorted(new Comparator<Clazz>() {
#Override
public int compare(Clazz o1, Clazz o2) {
return o1.getId().compareTo(o2.getId());
}
}).collect(Collectors.toList()); // Returns the sorted List (using toSet() wont work)

Elements in HashSet can't be sorted. Whenever you put elements into HashSet, it can mess up the ordering of the whole set. It is deliberately designed like that for performance. When you don't care about the order, HashSet will be the most efficient set for frequent insertions and queries.
TreeSet is the alternative that you can use. When you iterate on the tree set, you will get sorted elements automatically.
But it will adjust the tree to try to remain sorted every time you insert an element.
Perhaps, what you are trying to do is to sort just once. In that case, TreeSet is not the most efficient option because it needs to determine the placing of newly added elements all the time. Use TreeSet only when you want to sort often.
If you only need to sort once, use ArrayList. Create a new list and add all the elements then sort it once. If you want to retain only unique elements (remove all duplicates), then put the list into a LinkedHashSet, it will retain the order you have already sorted.
List<Integer> list = new ArrayList<>();
list.add(6);
list.add(4);
list.add(4);
list.add(5);
Collections.sort(list);
Set<Integer> unique = new LinkedHashSet<>(list); // 4 5 6
Now, you've gotten a sorted set if you want it in a list form then convert it into list.

You can use TreeSet as mentioned in other answers.
Here's a little more elaboration on how to use it:
TreeSet<String> ts = new TreeSet<String>();
ts.add("b1");
ts.add("b3");
ts.add("b2");
ts.add("a1");
ts.add("a2");
System.out.println(ts);
for (String s: ts)
System.out.println(s);
Output:
[a1, a2, a3, a4, a5]
a1
a2
b1
b2
b3

In my humble opinion , LazerBanana's answer should be the top rated answer & accepted because all the other answers pointing to java.util.TreeSet ( or first convert to list then call Collections.sort(...) on the converted list ) didn't bothered to ask OP as what kind of objects your HashSet has i.e. if those elements have a predefined natural ordering or not & that is not optional question but a mandatory question.
You just can't go in & start putting your HashSet elements into a TreeSet if element type doesn't already implement Comparable interface or if you are not explicitly passing Comparator to TreeSet constructor.
From TreeSet JavaDoc ,
Constructs a new, empty tree set, sorted according to the natural
ordering of its elements. All elements inserted into the set must
implement the Comparable interface. Furthermore, all such elements
must be mutually comparable: e1.compareTo(e2) must not throw a
ClassCastException for any elements e1 and e2 in the set. If the user
attempts to add an element to the set that violates this constraint
(for example, the user attempts to add a string element to a set whose
elements are integers), the add call will throw a ClassCastException.
That is why only all Java8 stream based answers - where you define your comparator on the spot - only make sense because implementing comparable in POJO becomes optional. Programmer defines comparator as and when needed. Trying to collect into TreeSet without asking this fundamental question is also incorrect ( Ninja's answer). Assuming object types to be String or Integer is also incorrect.
Having said that, other concerns like ,
Sorting Performance
Memory Foot Print ( retaining original set and creating new sorted sets each time sorting is done or wish to sort the set in - place etc etc )
should be the other relevant points too. Just pointing to API shouldn't be only intention.
Since Original set already contains only unique elements & that constraint is also maintained by sorted set so original set needs to be cleared from memory since data is duplicated.

1. Add all set element in list -> al.addAll(s);
2. Sort all the elements in list using -> Collections.sort(al);
public class SortSetProblem {
public static void main(String[] args) {
ArrayList<String> al = new ArrayList();
Set<String> s = new HashSet<>();
s.add("ved");
s.add("prakash");
s.add("sharma");
s.add("apple");
s.add("ved");
s.add("banana");
System.out.println("Before Sorting");
for (String s1 : s) {
System.out.print(" " + s1);
}
System.out.println("After Sorting");
al.addAll(s);
Collections.sort(al);
for (String set : al) {
System.out.print(" " + set);
}
}
}
input - ved prakash sharma apple ved banana
Output - apple banana prakash sharma ved

If you want want the end Collection to be in the form of Set and if you want to define your own natural order rather than that of TreeSet then -
Convert the HashSet into List
Custom sort the List using Comparator
Convert back the List into LinkedHashSet to maintain order
Display the LinkedHashSet
Sample program -
package demo31;
import java.util.*;
public class App26 {
public static void main(String[] args) {
Set<String> set = new HashSet<>();
addElements(set);
List<String> list = new LinkedList<>();
list = convertToList(set);
Collections.sort(list, new Comparator<String>() {
#Override
public int compare(String s1, String s2) {
int flag = s2.length() - s1.length();
if(flag != 0) {
return flag;
} else {
return -s1.compareTo(s2);
}
}
});
Set<String> set2 = new LinkedHashSet<>();
set2 = convertToSet(list);
displayElements(set2);
}
public static void addElements(Set<String> set) {
set.add("Hippopotamus");
set.add("Rhinocerous");
set.add("Zebra");
set.add("Tiger");
set.add("Giraffe");
set.add("Cheetah");
set.add("Wolf");
set.add("Fox");
set.add("Dog");
set.add("Cat");
}
public static List<String> convertToList(Set<String> set) {
List<String> list = new LinkedList<>();
for(String element: set) {
list.add(element);
}
return list;
}
public static Set<String> convertToSet(List<String> list) {
Set<String> set = new LinkedHashSet<>();
for(String element: list) {
set.add(element);
}
return set;
}
public static void displayElements(Set<String> set) {
System.out.println(set);
}
}
Output -
[Hippopotamus, Rhinocerous, Giraffe, Cheetah, Zebra, Tiger, Wolf, Fox, Dog, Cat]
Here the collection has been sorted as -
First - Descending order of String length
Second - Descending order of String alphabetical hierarchy

you can do this in the following ways:
Method 1:
Create a list and store all the hashset values into it
sort the list using Collections.sort()
Store the list back into LinkedHashSet as it preserves the insertion order
Method 2:
Create a treeSet and store all the values into it.
Method 2 is more preferable because the other method consumes lot of time to transfer data back and forth between hashset and list.

We can not decide that the elements of a HashSet would be sorted automatically. But we can sort them by converting into TreeSet or any List like ArrayList or LinkedList etc.
// Create a TreeSet object of class E
TreeSet<E> ts = new TreeSet<E> ();
// Convert your HashSet into TreeSet
ts.addAll(yourHashSet);
System.out.println(ts.toString() + "\t Sorted Automatically");

You can use guava library for the same
Set<String> sortedSet = FluentIterable.from(myHashSet).toSortedSet(new Comparator<String>() {
#Override
public int compare(String s1, String s2) {
// descending order of relevance
//required code
}
});

SortedSet has been added Since java 7
https://docs.oracle.com/javase/8/docs/api/java/util/SortedSet.html

You can wrap it in a TreeSet like this:
Set mySet = new HashSet();
mySet.add(4);
mySet.add(5);
mySet.add(3);
mySet.add(1);
System.out.println("mySet items "+ mySet);
TreeSet treeSet = new TreeSet(mySet);
System.out.println("treeSet items "+ treeSet);
output :
mySet items [1, 3, 4, 5]
treeSet items [1, 3, 4, 5]
Set mySet = new HashSet();
mySet.add("five");
mySet.add("elf");
mySet.add("four");
mySet.add("six");
mySet.add("two");
System.out.println("mySet items "+ mySet);
TreeSet treeSet = new TreeSet(mySet);
System.out.println("treeSet items "+ treeSet);
output:
mySet items [six, four, five, two, elf]
treeSet items [elf, five, four, six, two]
requirement for this method is that the objects of the set/list should be comparable (implement the Comparable interface)

The below is my sample code and its already answered by pointing the code in comments , am still sharing because it contains the complete code
package Collections;
import java.util.*;
public class TestSet {
public static void main(String[] args) {
Set<String> objset = new HashSet<>();
objset.add("test");
objset.add("abc");
objset.add("abc");
objset.add("mas");
objset.add("vas");
Iterator itset = objset.iterator();
while(itset.hasNext())
{
System.out.println(itset.next());
}
TreeSet<String> treeobj = new TreeSet(objset);
System.out.println(treeobj);
}
}
TreeSet treeobj = new TreeSet(objset); here we are invoking the treeset constructor which will call the addAll method to add the objects .
See this below code from the TreeSet class how its mentioned ,
public TreeSet(Collection<? extends E> c) {
this();
addAll(c);
}

Convert HashSet to List then sort it using Collection.sort()
List<String> list = new ArrayList<String>(hset);
Collections.sort(List)

This simple command did the trick for me:
myHashSet.toList.sorted
I used this within a print statement, so if you need to actually persist the ordering, you may need to use TreeSets or other structures proposed on this thread.

How to get element position from Java Map

I have this Java Map:
Can you tell me how I can get the 6-th element of the Map?
private static final Map<String, Users> cache = new HashMap<>();
is this possible? Or I have to use another Java collection?

Though a bit late to answer. But the option is to use LinkedHashMap: this map preserves the order according to insertion of elements, as everyone has suggested. However, As a warning, it has a constructor LinkedHashMap(int initialCapacity, float loadFactor, boolean accessOrder) which will create a linked hash map whose order of iteration is the order in which its entries were last accessed. Don't use this constructor for this case.
However, if I needed such functionality, i would extend it and implement my necessary function to re-use them in OOP way.
class MyLinkedMap<K, V> extends LinkedHashMap<K, V>
{
public V getValue(int i)
{
Map.Entry<K, V>entry = this.getEntry(i);
if(entry == null) return null;
return entry.getValue();
}
public Map.Entry<K, V> getEntry(int i)
{
// check if negetive index provided
Set<Map.Entry<K,V>>entries = entrySet();
int j = 0;
for(Map.Entry<K, V>entry : entries)
if(j++ == i)return entry;
return null;
}
}
Now i can instantiate it and can get a entry and value either way i want:
MyLinkedMap<String, Integer>map = new MyLinkedMap<>();
map.put("a first", 1);
map.put("a second", 2);
map.put("a third", 3);
System.out.println(map.getValue(2));
System.out.println(map.getEntry(1));
Output:
3
a second=2

HashMap doesn't grantee the order. If you concern about order you should use LinkedHashMap
Map<String, Users> orderedMap=new LinkedHashMap<>();
Now when you put an element it will keep the order what you put.
If you want to get 6th element, now you can do it since you have your elements in order.
orderedMap.values().toArray()[5]// will give you 6th value in the map.
Example
Map<String, String> orderedMap=new LinkedHashMap<>();
orderedMap.put("a","a");
orderedMap.put("b","b");
System.out.println(orderedMap.values().toArray()[1]); // you will get b(value)
System.out.println(orderedMap.keySet().toArray()[1]); // you will get b(key)
}

A HashMap does not maintain the order of the elements inserted in it. You can used a LinkedHashMap instead which maintains the order of the elements inserted in it.
Though you need to note that even a LinkedHashMap has no such method which would give the element at a particular index. You will have to manually iterate through the entries and extract the element at the 6th iteration.

With guava's Iterables
Iterables.get(map.entrySet(),6);

The HashMap has no defined ordering of keys.It's Unordered.
You can use LinkedHashMap which will store your keys in order of insertion.You can retrieve them by calling keySet().

HashMaps do not preserve ordering:
LinkedHashMap which guarantees a predictable iteration order.
Example
public class Users
{
private String Id;
public String getId()
{
return Id;
}
public void setId(String id)
{
Id = id;
}
}
Users user;
LinkedHashMap<String,Users> linkedHashMap = new LinkedHashMap<String,Users>();
for (int i = 0; i < 3; i++)
{
user = new Users();
user.setId("value"+i);
linkedHashMap.put("key"+i,user);
}
/* Get by position */
int pos = 1;
Users value = (new ArrayList<Users>(linkedHashMap.values())).get(pos);
System.out.println(value.getId());

According to documentation, HashMap is a Hash table based implementation of the Map interface. This implementation provides all of the optional map operations, and permits null values and the null key. (The HashMap class is roughly equivalent to Hashtable, except that it is unsynchronized and permits nulls.) This class makes no guarantees as to the order of the map; in particular, it does not guarantee that the order will remain constant over time.
That's why it is not wise to use this kind of Collection.
UPDATE:
Based on #Prateek implementation of LinkedHashMap I would suggest something like:
LinkedHashMap<String,User> linkedHashMap = new LinkedHashMap<String,User>();
// or LinkedHashMap<String,User> linkedHashMap = new LinkedHashMap<>(); //for java 7+
linkedHashMap.put("1",userObj1);
linkedHashMap.put("2",userObj2);
linkedHashMap.put("3",userObj3);
/* Get by position */
int pos = 1; // Your position
User tmp= (new ArrayList<User>(linkedHashMap.values())).get(pos);
System.out.println(tmp.getName());

A HashMap doesn't have a position. You can iterate through its KeySet or EntrySet, and pick the nth element, but it's not really the same as a position. A LinkedHashMap does have a position, since it has a predictable iteration order.

You need to use a LinkedHashMap in order to be able to tell the order of the inserted elements. HashMap is not capable of doing so.

There is no Order in HashMap. You can obtain the list of may keys using map.keySet() but there's no guarantee the key set will be in the order which you add it in. Use LinkedHashMap instead of HashMap It will always return keys in same order (as insertion)

Correct!!
you will have to use other collection for getting values on index(position).
You can use ArrayList

If the ordering is to mean anything significant, you could consider using a SortedMap.
Concrete implementation: TreeMap

Use LinkedHashMap instead of HashMap It will return keys in same order (as insertion) when calling keySet().
For mare detail about LinkedHashMap see this
For example to get the element from specific index
Create a new list from your values and get the value based on index.
LinkedHashMap<String, List<String>> hMap;
List<List<String>> l = new ArrayList<List<String>>(hMap.values());
l.get(6);

java beginner : How key gets sorted in hashmaps?

i am new to java and was learning the concepts of hashmaps.
I am confused how the keys are sorted in hashmaps.
i understood that its based on string length.
but i am confused how data is sorted when the string length is same.
import java.util.HashMap;
import java.util.Iterator;
import java.util.Map;
import java.util.Set;
public class HashMapExample
{
public static void main(String args[])
{
Map<String,String> map = new HashMap<String,String>(20);//SPECIFYING THE TYPE FOR FINDING HASH CODES.
//Adding values to the HashMap
map.put("key value a", "test value 1");
map.put("key value b", "test value 2");
map.put("key value c", "test value 3");
System.out.println("Retrieving values from HashMap");
retrieveValuesFromListMethod(map);
System.out.println("**********************");
}
/*This method retrieves values from Map
*/
public static void retrieveValuesFromListMethod(Map map)
{
Set keys = map.keySet();
Iterator itr = keys.iterator();
String key;
String value;
while(itr.hasNext())
{
key = (String)itr.next();
value = (String)map.get(key);
System.out.println(key + " - "+ value);
}
}
}
this is my code.
output is
Retrieving values from HashMap
key value c- test value 3
key value b- test value 2
key value a- test value 1
**********************
but instead of a,b,c if i give aa,ab,ac the output is different
Retrieving values from HashMap
key value ab - test value 2
key value aa - test value 1
key value ac - test value 3
**********************
for 1,2,3
Retrieving values from HashMap
key value 1 - test value 1
key value 2 - test value 2
key value 3 - test value 3
**********************
how sorting is done in hashmap? Please help!!
Thanks in advance.

java.util.HashMap is unordered; you can't and shouldn't assume
anything beyond that.
This class makes no guarantees as to the order of the map; in particular, it does not guarantee that the order will remain constant
over time.
java.util.LinkedHashMap uses insertion-order.
This implementation differs from HashMap in that it maintains a doubly-linked list running through all of its entries. This linked
list defines the iteration ordering, which is normally the order in
which keys were inserted into the map (insertion-order).
java.util.TreeMap, a SortedMap, uses either natural or custom ordering
of the keys.
The map is sorted according to the natural ordering of its keys, or by a Comparator provided at map creation time, depending on which
constructor is used.

java beginner : How key gets sorted in hashmaps?
Taking from answer from here which is also the answer to your question
Use sorted TreeMap:
Map<String, Float> map = new TreeMap<String, Float>(yourMap);
It will automatically put entries sorted by keys. I think natural String ordering will be fine in your case.
Note that HashMap due to lookup optimizations does not preserve order.
If you want to preserve the order in which the data was inserted into the map you can use LinkedHashMap.

In your case , when you are using the following
//Adding values to the HashMap
map.put("key value a", "test value 1");
map.put("key value b", "test value 2");
map.put("key value c", "test value 3");
here is the snap of hashmap after the insertion of the keys
HashMap internally uses Array of Entry Map , hashcode generated by the keys are input to the hashing function that makes this key to be inserted in the array in the order you are looking into it .
You are using iterator to iterate the hashmap ,have a look at the snap of iterator for the above collection, since the iterator is looking at the collection in the following order , it just seems the keys are sorted but are actually not .
(Note : Iterator doesn't guarantee for the iteration in the order in which the elements are present in collection )
so its just a resemblance not the actual behavior of hashmap . This behavior for sorted key is given by TreeMap or any collection implementing interface SortedMap and the key when comparable .
I hope above information will be helpful to you

A Hashmap is not sorted.
If you have to keep the order, you have to use e.g. LinkedHashMap<K, V>

Try TreeMap,
Map<String, String> sampleMap = new TreeMap<String, String>(map);
use TreeMap the all keys are in sorted order

hashmap is not sorted, the output is different because the String's hashcode is not same.

Best way to create a hashmap of arraylist

I have one million rows of data in .txt format. the format is very simple. For each row:
user1,value1
user2,value2
user3,value3
user1,value4
...
You know what I mean. For each user, it could appear many times, or appear only once (you never know). I need to find out all the values for each user. Because user may appear randomly, I used Hashmap to do it. That is: HashMap(key: String, value: ArrayList). But to add data to the arrayList, I have to constantly use HashMap get(key) to get the arrayList, add value to it, then put it back to HashMap. I feel it is not that very efficient. Anybody knows a better way to do that?

You don't need to re-add the ArrayList back to your Map. If the ArrayList already exists then just add your value to it.
An improved implementation might look like:
Map<String, Collection<String>> map = new HashMap<String, Collection<String>>();
while processing each line:
String user = user field from line
String value = value field from line
Collection<String> values = map.get(user);
if (values==null) {
values = new ArrayList<String>();
map.put(user, values)
}
values.add(value);
Follow-up April 2014 - I wrote the original answer back in 2009 when my knowledge of Google Guava was limited. In light of all that Google Guava does, I now recommend using its Multimap instead of reinvent it.
Multimap<String, String> values = HashMultimap.create();
values.put("user1", "value1");
values.put("user2", "value2");
values.put("user3", "value3");
values.put("user1", "value4");
System.out.println(values.get("user1"));
System.out.println(values.get("user2"));
System.out.println(values.get("user3"));
Outputs:
[value4, value1]
[value2]
[value3]

Use Multimap from Google Collections. It allows multiple values for the same key
https://google.github.io/guava/releases/19.0/api/docs/com/google/common/collect/Multimap.html

Since Java 8 you can use map.computeIfAbsent
https://docs.oracle.com/javase/8/docs/api/java/util/Map.html#computeIfAbsent-K-java.util.function.Function-
Collection<String> values = map.computeIfAbsent(user, k -> new ArrayList<>());
values.add(value);

The ArrayList values in your HashMap are references. You don't need to "put it back to HashMap". You're operating on the object that already exists as a value in the HashMap.

If you don't want to import a library.
package util;
import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
/**
* A simple implementation of a MultiMap. This implementation allows duplicate elements in the the
* values. (I know classes like this are out there but the ones available to me didn't work).
*/
public class MultiMap<K, V> extends HashMap<K, List<V>> {
/**
* Looks for a list that is mapped to the given key. If there is not one then a new one is created
* mapped and has the value added to it.
*
* #param key
* #param value
* #return true if the list has already been created, false if a new list is created.
*/
public boolean putOne(K key, V value) {
if (this.containsKey(key)) {
this.get(key).add(value);
return true;
} else {
List<V> values = new ArrayList<>();
values.add(value);
this.put(key, values);
return false;
}
}
}

i think what you want is the Multimap. You can get it from apache's commons collection, or google-collections.
http://commons.apache.org/collections/
http://code.google.com/p/google-collections/
"collection similar to a Map, but
which may associate multiple values
with a single key. If you call put(K,
V) twice, with the same key but
different values, the multimap
contains mappings from the key to both
values."

I Could not find any easy way. MultiMap is not always an option available. So I wrote something this.
public class Context<K, V> extends HashMap<K, V> {
public V addMulti(K paramK, V paramV) {
V value = get(paramK);
if (value == null) {
List<V> list = new ArrayList<V>();
list.add(paramV);
put(paramK, paramV);
} else if (value instanceof List<?>) {
((List<V>)value).add(paramV);
} else {
List<V> list = new ArrayList<V>();
list.add(value);
list.add(paramV);
put(paramK, (V) list);
}
return paramV;
}
}

it would be faster if you used a LinkedList instead of an ArrayList, as the ArrayList will need to resize when it nears capacity.
you will also want to appropriately estimate the capacity of the wrapping collection (HashMap or Multimap) you are creating to avoid repetitive rehashing.

As already mentioned, MultiMap is your best option.
Depending on your business requirements or constraints on the data file, you may want to consider doing a one-off sorting of it, to make it more optimised for loading.