Is there anyway I can print out the Keys with the values that had collisions ? If a key has two values I want to be able to print out those values how can I do that I am inside the hash map class and making modification to it.
In a HashMap, a key cannot have two values. If you call map.put(key,value) with an existing key, the old value is removed from the map, and is returned by put().
One way to have multiple values per key is by using HashMap<K,Collection<V>>. This automatically provides the functionality you want, since you can simply examine the contents of the value collection after you've added the new element to it.
There are also third-party classes that provide this functionality, such as MultiValueMap.
edit:
If you're talking about multiple keys ending up in the same bucket, then you need to modify HashMap's put() method:
public V put(K key, V value) { if (key == null) return putForNullKey(value); int hash = hash(key.hashCode()); int i = indexFor(hash, table.length); if (table[i] != null) { // TODO: there's already something in this bucket } for (Entry<K,V> e = table[i]; e != null; e = e.next) { ...
(Add your code where the TODO line is.)
You'll need to make similar changes to putForNullKey() and other related methods, such as putForCreate().
If you're modifying the HashMap class then you should be able to find where the list of elements is chained off the hash array and detect when there's more than one in the chain.
HOWEVER, you'd better have a very good reason for modifying HashMap, AND you'd better change the package and name of the class (to something like com.my.company.HashMapWithCollisionStatistics) or risk the eternal wrath of everybody who comes behind you to maintain Java code in your group.
Why you need to do that? If just want to check the collision then call hashcode() on the objects that you are trying to store in the hashmap without storing them.
I have a 2.67 GHz Celeron processor, 1.21 GB of RAM on a x86 Windows XP Professional machine. My understanding is that the Android emulator should start fairly quickly on such a machine, but for me it does not. I have followed all instructions in setting up the IDE, SDKs, JDKs and such and have had some success in staring the emulator quickly but is very particulary. How can I, if possible, fix this problem?
If a key has two values...
ReplyDeleteIn a HashMap, a key cannot have two values. If you call map.put(key,value) with an existing key, the old value is removed from the map, and is returned by put().
One way to have multiple values per key is by using HashMap<K,Collection<V>>. This automatically provides the functionality you want, since you can simply examine the contents of the value collection after you've added the new element to it.
There are also third-party classes that provide this functionality, such as MultiValueMap.
edit:
If you're talking about multiple keys ending up in the same bucket, then you need to modify HashMap's put() method:
public V put(K key, V value) {
if (key == null)
return putForNullKey(value);
int hash = hash(key.hashCode());
int i = indexFor(hash, table.length);
if (table[i] != null) {
// TODO: there's already something in this bucket
}
for (Entry<K,V> e = table[i]; e != null; e = e.next) {
...
(Add your code where the TODO line is.)
You'll need to make similar changes to putForNullKey() and other related methods, such as putForCreate().
If you're modifying the HashMap class then you should be able to find where the list of elements is chained off the hash array and detect when there's more than one in the chain.
ReplyDeleteHOWEVER, you'd better have a very good reason for modifying HashMap, AND you'd better change the package and name of the class (to something like com.my.company.HashMapWithCollisionStatistics) or risk the eternal wrath of everybody who comes behind you to maintain Java code in your group.
Just use MultiMap from google collections.
ReplyDeleteWhy you need to do that? If just want to check the collision then call hashcode() on the objects that you are trying to store in the hashmap without storing them.
ReplyDelete