Convert an int to a letter value

Question

Out of interest I have been working on a very basic encoder/decoder program. Super simple encoding, a = 1, b = 2... and from there it prints out to the numbers : "1 2 3". I have that working:

    final Map<Character, Integer> map;
    final String str = "hello world";

    map = new HashMap<>();  
    before Java 7.
    map.put('a', 1);
    map.put('b', 2);
    map.put('c', 3);
    map.put('d', 4);
    map.put('e', 5);
    map.put('f', 6);
    map.put('g', 7);
    map.put('h', 8);
    map.put('i', 9);
    map.put('j', 10);
    map.put('k', 11);
    map.put('l', 12);
    map.put('m', 13);
    map.put('n', 14);
    map.put('o', 15);
    map.put('p', 16);
    map.put('q', 17);
    map.put('r', 18);
    map.put('s', 19);
    map.put('t', 20);
    map.put('u', 21);
    map.put('v', 22);
    map.put('w', 23);
    map.put('x', 24);
    map.put('y', 25);
    map.put('z', 26);

    for(final char c : str.toCharArray())
    {
        final Integer val;

        val = map.get(c);

        if(val == null)
        {   
            //error
        }
        else
        {
            System.out.print(val + " ");
        }
    }

    System.out.println();
}

Now I am working on a way to convert it back. I think I could do it with a reverse hashmap of sorts but I can't get it to work. Any suggestions? Thanks.

Answer 1

You don't need a HashMap for this data at all. Take advantage of the fact that the character values for these letters are all consecutive numbers. That is, 'a' is 97, 'b' is 98, ..., 'z' is 122.

You can map a letter character to a number by subtracting96.

val = c - 96;

You can map the number back to the letter by adding 96, then casting it back to a char.

c = (char) (val + 96); 

Answer 2

From the documentation, you can check to see whether the value has a key pre-image and what key it is. In fact, there are two functions that help this cause:

boolean containsValue(Object value)

Returns true if this map maps one or more keys to the specified value. More formally, returns true if and only if this map contains at least one mapping to a value v such that (value==null ? v==null : value.equals(v)). This operation will probably require time linear in the map size for most implementations of the Map interface

Second method

K getKey(Object value)

Gets the key that is currently mapped to the specified value.

If the value is not contained in the map, null is returned.

Answer 3

Something like this should do it:

Integer integerToFind = ...;
Character matchingKey = null;
for (Entry<Character, Integer> entry : map.entrySet()) {
    if (entry.getValue().equals(integerToFind)) {
        matchingKey = entry.getKey();
        break;
    }
}

This assumes there is no keys with the same value.

Answer 4

I like rgettman's answer but if you ever want a non ordered code this way might work better.

This uses Guava's BiMap https://code.google.com/p/guava-libraries/wiki/NewCollectionTypesExplained#BiMap

import com.google.common.collect.BiMap;
import com.google.common.collect.HashBiMap;


public class Encoder
{

    private BiMap<Character, Integer> mCodeMap = makeCodeMap();

    public String encode(String string) 
    {
        string = string.trim();
        if (string.length() <= 0) return "";

        final StringBuilder results = new StringBuilder();
        char c = string.charAt(0);
        results.append(mCodeMap.get(c));

        for (int i = 1; i < string.length(); ++i) {
            c = string.charAt(i);
            results.append(' ').append(mCodeMap.get(c));
        }

        return results.toString();
    }

    public String decode(String string)
    {
        string = string.trim();
        if (string.length() <= 0) return "";

        String[] nums = string.split(" ");

        final StringBuilder results = new StringBuilder();

        final BiMap<Integer, Character>inverse = mCodeMap.inverse();

        for (int i = 0; i < nums.length; ++i) {
            final int num = Integer.parseInt(nums[i]);
            results.append(inverse.get(num));
        }

        return results.toString();
    }

    private static BiMap<Character, Integer> makeCodeMap()
    {
        BiMap<Character, Integer> codeMap = HashBiMap.create();

        codeMap.put('a', 1);
        codeMap.put('b', 2);
        codeMap.put('c', 3);
        codeMap.put('d', 4);
        codeMap.put('e', 5);
        codeMap.put('f', 6);
        codeMap.put('g', 7);
        codeMap.put('h', 8);
        codeMap.put('i', 9);
        codeMap.put('j', 10);
        codeMap.put('k', 11);
        codeMap.put('l', 12);
        codeMap.put('m', 13);
        codeMap.put('n', 14);
        codeMap.put('o', 15);
        codeMap.put('p', 16);
        codeMap.put('q', 17);
        codeMap.put('r', 18);
        codeMap.put('s', 19);
        codeMap.put('t', 20);
        codeMap.put('u', 21);
        codeMap.put('v', 22);
        codeMap.put('w', 23);
        codeMap.put('x', 24);
        codeMap.put('y', 25);
        codeMap.put('z', 26);
        codeMap.put(' ', 27);

        return codeMap;
    }

    public static void main(String[] args)
    {       
        final String inputString = "this is my test input string";

        System.out.println("String = " + inputString);

        Encoder encoder = new Encoder();

        final String encoded = encoder.encode(inputString);
        System.out.println("Encoded = " + encoded);

        final String decoded = encoder.decode(encoded);
        System.out.println("Decoded = " + decoded);

        System.out.println("Correct = " + inputString.equals(decoded));
    }

}