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Can I declare a HashMap instead of a HashMap?
Differences between HashMap and Hashtable?Is Java “pass-by-reference” or “pass-by-value”?Sort a Map<Key, Value> by valuesHow can I create an executable JAR with dependencies using Maven?How to get an enum value from a string value in Java?Iterate through a HashMapHow do I determine whether an array contains a particular value in Java?How do I declare and initialize an array in Java?Java Hashmap: How to get key from value?How to update a value, given a key in a java hashmap?
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I'm trying to insert keys into a hashmap but I don't really need values inserted. I could but they wouldn't be used. I know hashmaps can accept null keys and values but only one null key,value pair. I could map.put(key,null) with the values being null but that method seems inefficient. My intention is that I'm going to use the map.containsKey(key) method to determine if a key exists in a hashmap which is why I don't need a value.
With that being said, is there a way to declare HashMap<key> instead of HashMap<key,arbitraryValue> so I won't have to add unnecessary null values? Sorry if this may be a dumb question.
java hashmap
asked Mar 8 at 1:58
AlexAlex
31
add a comment |
I'm trying to insert keys into a hashmap but I don't really need values inserted. I could but they wouldn't be used. I know hashmaps can accept null keys and values but only one null key,value pair. I could map.put(key,null) with the values being null but that method seems inefficient. My intention is that I'm going to use the map.containsKey(key) method to determine if a key exists in a hashmap which is why I don't need a value.
With that being said, is there a way to declare HashMap<key> instead of HashMap<key,arbitraryValue> so I won't have to add unnecessary null values? Sorry if this may be a dumb question.
java hashmap
asked Mar 8 at 1:58
AlexAlex
31
9
The data type that you're looking for is called aSet<E>.
– Jacob G.
Mar 8 at 2:00
1
Why do you need aHashMapif you don't need key value pairs? Why not just use a list or a set?
– Jeremy
Mar 8 at 2:00
4
TheHashSetclass that you should be using, is implemented usingHashMap, so apprently it’s not so inefficient as one might think. Don’t worry about it.
– Ole V.V.
Mar 8 at 2:07
I think Google might have an implementation of the HashSet that you might find interesting.
– Adrian M.
Mar 8 at 3:18
add a comment |
I'm trying to insert keys into a hashmap but I don't really need values inserted. I could but they wouldn't be used. I know hashmaps can accept null keys and values but only one null key,value pair. I could map.put(key,null) with the values being null but that method seems inefficient. My intention is that I'm going to use the map.containsKey(key) method to determine if a key exists in a hashmap which is why I don't need a value.
With that being said, is there a way to declare HashMap<key> instead of HashMap<key,arbitraryValue> so I won't have to add unnecessary null values? Sorry if this may be a dumb question.
java hashmap
asked Mar 8 at 1:58
AlexAlex
31
I'm trying to insert keys into a hashmap but I don't really need values inserted. I could but they wouldn't be used. I know hashmaps can accept null keys and values but only one null key,value pair. I could map.put(key,null) with the values being null but that method seems inefficient. My intention is that I'm going to use the map.containsKey(key) method to determine if a key exists in a hashmap which is why I don't need a value.
With that being said, is there a way to declare HashMap<key> instead of HashMap<key,arbitraryValue> so I won't have to add unnecessary null values? Sorry if this may be a dumb question.
java hashmap
java hashmap
asked Mar 8 at 1:58
AlexAlex
31
asked Mar 8 at 1:58
AlexAlex
31
asked Mar 8 at 1:58
AlexAlex
31
asked Mar 8 at 1:58
AlexAlex
31
asked Mar 8 at 1:58
AlexAlex
31
31
9
The data type that you're looking for is called aSet<E>.
– Jacob G.
Mar 8 at 2:00
1
Why do you need aHashMapif you don't need key value pairs? Why not just use a list or a set?
– Jeremy
Mar 8 at 2:00
4
TheHashSetclass that you should be using, is implemented usingHashMap, so apprently it’s not so inefficient as one might think. Don’t worry about it.
– Ole V.V.
Mar 8 at 2:07
I think Google might have an implementation of the HashSet that you might find interesting.
– Adrian M.
Mar 8 at 3:18
add a comment |
9
The data type that you're looking for is called aSet<E>.
– Jacob G.
Mar 8 at 2:00
1
Why do you need aHashMapif you don't need key value pairs? Why not just use a list or a set?
– Jeremy
Mar 8 at 2:00
4
TheHashSetclass that you should be using, is implemented usingHashMap, so apprently it’s not so inefficient as one might think. Don’t worry about it.
– Ole V.V.
Mar 8 at 2:07
I think Google might have an implementation of the HashSet that you might find interesting.
– Adrian M.
Mar 8 at 3:18
9
9
The data type that you're looking for is called a
Set<E>.– Jacob G.
Mar 8 at 2:00
The data type that you're looking for is called a
Set<E>.– Jacob G.
Mar 8 at 2:00
1
1
Why do you need a
HashMap if you don't need key value pairs? Why not just use a list or a set?– Jeremy
Mar 8 at 2:00
Why do you need a
HashMap if you don't need key value pairs? Why not just use a list or a set?– Jeremy
Mar 8 at 2:00
4
4
The
HashSet class that you should be using, is implemented using HashMap, so apprently it’s not so inefficient as one might think. Don’t worry about it.– Ole V.V.
Mar 8 at 2:07
The
HashSet class that you should be using, is implemented using HashMap, so apprently it’s not so inefficient as one might think. Don’t worry about it.– Ole V.V.
Mar 8 at 2:07
I think Google might have an implementation of the HashSet that you might find interesting.
– Adrian M.
Mar 8 at 3:18
I think Google might have an implementation of the HashSet that you might find interesting.
– Adrian M.
Mar 8 at 3:18
add a comment |
1 Answer
1
active
oldest
votes
As Jacob G. said, you should use a Set<E>.
A set is just a collection that does not contain any duplicates. The implementing class HashSet<E> actually uses a HashMap<K,V> under the hood, as mentioned by Ole V.V., but using a Set<E> in your code is the better approach IMO, because your problem does not require the values.
Additionally, there is a problem with using a HashMap<K,V> where all your values are null is in the get(K key) method. This method will return null if the requested key does not have an associated value. So how do you know if your call to get returned a valid or invalid null? i.e.
Map<Integer, Object> map = new HashMap<>();
// add entry 0 ->
map.put(0, null);
Object get1 = map.get(0); // returns null, so 0 must be a key in our map!
Object get2 = map.get(1); // also returns null, so is 1 a key too? No!
So, for your specific problem, I would try something like this!
Set<MyKey> set = new HashSet<>(); // or any implementing class
...
MyKey someKey = ...
// Check if your key set doesn't have some key, if so add it to the key set
if (!set.contains(someKey))
set.add(someKey);
answered Mar 8 at 3:20
Derek PlautzDerek Plautz
5112
add a comment |
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1 Answer
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As Jacob G. said, you should use a Set<E>.
A set is just a collection that does not contain any duplicates. The implementing class HashSet<E> actually uses a HashMap<K,V> under the hood, as mentioned by Ole V.V., but using a Set<E> in your code is the better approach IMO, because your problem does not require the values.
Additionally, there is a problem with using a HashMap<K,V> where all your values are null is in the get(K key) method. This method will return null if the requested key does not have an associated value. So how do you know if your call to get returned a valid or invalid null? i.e.
Map<Integer, Object> map = new HashMap<>();
// add entry 0 ->
map.put(0, null);
Object get1 = map.get(0); // returns null, so 0 must be a key in our map!
Object get2 = map.get(1); // also returns null, so is 1 a key too? No!
So, for your specific problem, I would try something like this!
Set<MyKey> set = new HashSet<>(); // or any implementing class
...
MyKey someKey = ...
// Check if your key set doesn't have some key, if so add it to the key set
if (!set.contains(someKey))
set.add(someKey);
answered Mar 8 at 3:20
Derek PlautzDerek Plautz
5112
add a comment |
As Jacob G. said, you should use a Set<E>.
A set is just a collection that does not contain any duplicates. The implementing class HashSet<E> actually uses a HashMap<K,V> under the hood, as mentioned by Ole V.V., but using a Set<E> in your code is the better approach IMO, because your problem does not require the values.
Additionally, there is a problem with using a HashMap<K,V> where all your values are null is in the get(K key) method. This method will return null if the requested key does not have an associated value. So how do you know if your call to get returned a valid or invalid null? i.e.
Map<Integer, Object> map = new HashMap<>();
// add entry 0 ->
map.put(0, null);
Object get1 = map.get(0); // returns null, so 0 must be a key in our map!
Object get2 = map.get(1); // also returns null, so is 1 a key too? No!
So, for your specific problem, I would try something like this!
Set<MyKey> set = new HashSet<>(); // or any implementing class
...
MyKey someKey = ...
// Check if your key set doesn't have some key, if so add it to the key set
if (!set.contains(someKey))
set.add(someKey);
answered Mar 8 at 3:20
Derek PlautzDerek Plautz
5112
add a comment |
As Jacob G. said, you should use a Set<E>.
A set is just a collection that does not contain any duplicates. The implementing class HashSet<E> actually uses a HashMap<K,V> under the hood, as mentioned by Ole V.V., but using a Set<E> in your code is the better approach IMO, because your problem does not require the values.
Additionally, there is a problem with using a HashMap<K,V> where all your values are null is in the get(K key) method. This method will return null if the requested key does not have an associated value. So how do you know if your call to get returned a valid or invalid null? i.e.
Map<Integer, Object> map = new HashMap<>();
// add entry 0 ->
map.put(0, null);
Object get1 = map.get(0); // returns null, so 0 must be a key in our map!
Object get2 = map.get(1); // also returns null, so is 1 a key too? No!
So, for your specific problem, I would try something like this!
Set<MyKey> set = new HashSet<>(); // or any implementing class
...
MyKey someKey = ...
// Check if your key set doesn't have some key, if so add it to the key set
if (!set.contains(someKey))
set.add(someKey);
answered Mar 8 at 3:20
Derek PlautzDerek Plautz
5112
As Jacob G. said, you should use a Set<E>.
A set is just a collection that does not contain any duplicates. The implementing class HashSet<E> actually uses a HashMap<K,V> under the hood, as mentioned by Ole V.V., but using a Set<E> in your code is the better approach IMO, because your problem does not require the values.
Additionally, there is a problem with using a HashMap<K,V> where all your values are null is in the get(K key) method. This method will return null if the requested key does not have an associated value. So how do you know if your call to get returned a valid or invalid null? i.e.
Map<Integer, Object> map = new HashMap<>();
// add entry 0 ->
map.put(0, null);
Object get1 = map.get(0); // returns null, so 0 must be a key in our map!
Object get2 = map.get(1); // also returns null, so is 1 a key too? No!
So, for your specific problem, I would try something like this!
Set<MyKey> set = new HashSet<>(); // or any implementing class
...
MyKey someKey = ...
// Check if your key set doesn't have some key, if so add it to the key set
if (!set.contains(someKey))
set.add(someKey);
answered Mar 8 at 3:20
Derek PlautzDerek Plautz
5112
answered Mar 8 at 3:20
Derek PlautzDerek Plautz
5112
answered Mar 8 at 3:20
Derek PlautzDerek Plautz
5112
answered Mar 8 at 3:20
Derek PlautzDerek Plautz
5112
5112
add a comment |
add a comment |
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9
The data type that you're looking for is called a
Set<E>.– Jacob G.
Mar 8 at 2:00
1
Why do you need a
HashMapif you don't need key value pairs? Why not just use a list or a set?– Jeremy
Mar 8 at 2:00
4
The
HashSetclass that you should be using, is implemented usingHashMap, so apprently it’s not so inefficient as one might think. Don’t worry about it.– Ole V.V.
Mar 8 at 2:07
I think Google might have an implementation of the HashSet that you might find interesting.
– Adrian M.
Mar 8 at 3:18