Java datastructure to hold a value mapped to an array

Question

I am kind of lost on which data structure to use to solve my problem effectively. I want to map an array to a value. What I mean is that if I have 1000 values, I need to be able to map multiple other values to each of the 1000 values.

For example,

I have 1000 A values from 1-1000. For each value A, I want to map k other values B (these range from 1-1000 also). But I do want to ensure that whatever values are mapped to A are not duplicates. Mapped values between different A values can be the same (i.e. both 2 and 1000 have 67 mapped to them).

    1 -> 138, 92, 835, 841, 12
    2 -> 766, 324, 26, 933, 62
    3 -> 53, 131, 62, 121, 67
    4->160, 160 #NOT OK
    4-> 162, 171, 594, 912, 455
    ...
    1000->146, 981, 67, 246, 146

So when I look at some arbitrary value A, I should be easily able to identify whatever values are mapped to it. So if I wanted to access value 3, I should be able to print out both the value A (3) and its associated values (53, 131, 62, 121, 67).

I hope that makes sense. What would be the best way of achieving this kind of data structure? Any help of an explanation or an example would be much appreciated.

Jigar Joshi · Accepted Answer · 2013-04-02 03:05:17Z

2

Use

Map<Long, Set<Long>> numberToValuesMap;

Set for uniqueness

answered Apr 2, 2013 at 3:05

Jigar Joshi

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2 Comments

PaulProgrammer Over a year ago

Above with map would work too, but the natural order of the keys may not be preserved -- you'd have to sort the key set before iterating if you needed to preserve order.

Jigar Joshi Over a year ago

You could use TreeSet to preserve the order

PaulProgrammer · Accepted Answer · 2013-04-02 03:51:17Z

1

You want an array list of sets:

ArrayList<Set<Long>>

Set requires that the value is in the collection once and only once, and of course the list is a list of these collections.

You can then use the get(index) method on ArrayList to get specific ordinals.

ArrayList<Set<Number>> mappings = new ArrayList<Set<Long>>();

Set<Long> s = new HashSet<Long>();
long[] n = {138, 92, 835, 841, 12};
s.addAll( Arrays.toList( n));
mappings.add(1, s);
// etc.

Later, to fetch:

Set<Long> result = mappings.get(1); // for element in slot 1...

edited Apr 2, 2013 at 3:51

answered Apr 2, 2013 at 3:10

PaulProgrammer

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2 Comments

Jigar Joshi Over a year ago

retrieval would be easier & efficient with Map

PaulProgrammer Over a year ago

Retrieval is just as easy using list (because the ordinal is essentially the map key) and has an O(1) lookup time, because it's backed by an array, which is not true of a Map.

Bohemian · Accepted Answer · 2013-04-02 03:16:05Z

0

if you have continuous values (or even near-continuous), you don't need to "map" then - you can use a simple array of array:

int[][] a = new int[1001][];

then for each value, create an array of int:

a[1] = new int[]{138, 92, 835, 841, 12};

etc

answered Apr 2, 2013 at 3:16

Bohemian♦

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2 Comments

PaulProgrammer Over a year ago

You'd have a harder time ensuring the "unique" constraint with this implementation.

Paul Over a year ago

That seems to be the only drawback. Otherwise it looks to be very simple and straightforward which I like.

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