CODEWITHSUNDEEP
javahashset
Pulkit
Pulkit

Reputation: 65

index of character in hashset Java

How do I get the index of element in HashSet when using java?

Let's say 

HashSet<Character> hs = new HashSet<Character>();
hs.add('s');
hs.add('t');
hs.add('a');
hs.add('c');

// some code

if (hs.contains('a')) {
   int a = // need the index of a here from hashset 
}

The runtime for lookup has to be linear with any data structure.

Upvotes: 0

Views: 9260

Answers (4)

Azhy
Azhy

Reputation: 706

HashSet Class hasn't any functions to find the index of something inside it, So you can use something like ArrayList or List, But if you want to use HashSet any way, Then only way to find the item's index is by iterating it like:-

HashSet<Character> hs = new HashSet<Character();
hs.add('s');
hs.add('t');
hs.add('a');
hs.add('c');

int foundIndex = -1;

for(int i=0; i<hs.size(); i++){
    if(hs.get(i).equals('a')){
        foundIndex = i;
        break;
    }
}

if(foundIndex == -1){
    //The value wasn't in the HashSet
}else{
    int a = foundIndex;
    //The value was found
}

Upvotes: 0

freakomonk
freakomonk

Reputation: 614

A Set in unordered, so you cannot get an index from it. If you still want to get index from a set, you can opt for iterating through it. 

HashSet<Character> hs = new HashSet<Character>();
Iterator it = hs.iterator();
int index =-1;
 while(it.hasNext()){
  i++;
  if(value.next()=='a'){
  return;
 }
}
return i== (hs.size()-1) ? -1 : i;

Constant time lookup will be traded off in this way because of unorderedness of HashSet.

Upvotes: 0

vojislavdjukic
vojislavdjukic

Reputation: 448

The set does not have the index by definition. Instead, try to use something like HashMap. This would give you constant time for containsKey:

HashMap<Character, Integer> hm = new HashMap<Character, Integer>();
hm.put('s',1);
hm.put('t',2);
hm.put('a',3);
hm.put('c',4);

// some code

if (hm.containsKey('a')){
int a = hm.get('a') //index is here

}

Upvotes: 0

PradyumanDixit
PradyumanDixit

Reputation: 2375

The Set interface doesn't have something like as an indexOf() method. You'd really need to iterate over it or to use the List interface instead which offers an indexOf() method.

If you would like to, converting Set to List is pretty simple, it should be a matter of passing the Set through the constructor of the List implementation. E.g.

List<String> nameList = new ArrayList<String>(nameSet);

Upvotes: 3

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