LetterPhoneNumber.java

package backtracking;

import java.util.ArrayList;
import java.util.List;

/**
 * Created by gouthamvidyapradhan on 09/03/2017.
 * Given a digit string, return all possible letter combinations that the number could represent.
 * <p>
 * A mapping of digit to letters (just like on the telephone buttons) is given below.
 * 1 2(abc) 3(def)
 * 4(ghi) 5(jkl) 6(mno)
 * 7(pqrs) 8(tuv) 9(wxyz)
 * <p>
 * <p>
 * Input:Digit string "23"
 * Output: ["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"].
 * Note:
 * Although the above answer is in lexicographical order, your answer could be in any order you want.
 */
public class LetterPhoneNumber {
    private String[] NUMBER_ALPHA = {"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};

    /**
     * Main method
     *
     * @param args
     * @throws Exception
     */
    public static void main(String[] args) throws Exception {
        List<String> result = new LetterPhoneNumber().letterCombinations("23");
        result.forEach(System.out::println);
    }

    private List<String> letterCombinations(String digits) {
        if (digits == null || digits.isEmpty() || digits.contains("1") || digits.contains("0"))
            return new ArrayList<>();
        List<String> prev = new ArrayList<>();
        prev.add("");
        for (int i = digits.length() - 1; i >= 0; i--) {
            String str = NUMBER_ALPHA[Integer.parseInt(String.valueOf(digits.charAt(i)))];
            List<String> newList = new ArrayList<>();
            for (int j = 0, l = str.length(); j < l; j++) {
                for (String s : prev) {
                    s = str.charAt(j) + s;
                    newList.add(s);
                }
            }
            prev = newList;
        }
        return prev;
    }


}