In Java please, it is ok to use stack or queue or string builder but no hash set

Question

In Java please, it is ok to use stack or queue or string builder but no hash set

. Write a recursive method that will find and list all of the one-element sequences of a letters in a char[] array, then all the two-element sequences, then all of the three ele- letters in a char[] array, then all the two-element sequences,then all of the three ele- ment sequences, and so on, such that the characters in each sequence appear in the r as they are in the array. For example, for the following arrays char[] letters = {'A', 'C', 'E', 'G'); the one-element sequences are "A", "C", "E", and "G" the two-element sequences are "AC", "AE", "AG", "CE", "CG", "EG" the three-element sequences are "ACE", "ACG", "AEG", "CEG" the four-element sequence is "ACEG"

Explanation / Answer

import java.util.ArrayList; import java.util.List; import muset.Alphabet.Letter; public class Sequence { private final List letters; public final Alphabet alphabet; public Sequence(Alphabet alphabet, List letters) { super(); this.letters = letters; this.alphabet = alphabet; } public Letter letterAt(int i) { return letters.get(i); } /** * * @param letter * @return A new Sequence containing all the letters in this plus an additional one at the end. */ public Sequence append(Letter letter) { List result = new ArrayList(letters); result.add(letter); return new Sequence(alphabet, result); } public int length() { return letters.size(); } @Override public String toString() { return letters.toString(); } @Override public int hashCode() { final int prime = 31; int result = 1; result = prime * result + ((letters == null) ? 0 : letters.hashCode()); return result; } @Override public boolean equals(Object obj) { if (this == obj) return true; if (obj == null) return false; if (getClass() != obj.getClass()) return false; Sequence other = (Sequence) obj; if (letters == null) { if (other.letters != null) return false; } else if (!letters.equals(other.letters)) return false; return true; } /** * * @param string * @return A sequence where each letter is on Character long */ public static Sequence buildSimpleSequence(Alphabet alphabet, String string) { List result = new ArrayList(string.length()); for (char c : string.toCharArray()) result.add(alphabet.getLetter("" + c)); return new Sequence(alphabet, result); } public Sequence subsequence(int fromIndex, int toIndex) { return new Sequence(alphabet, letters.subList(fromIndex, toIndex)); } }

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