17. Letter Combinations of a Phone Number

Given a digit string, return all possible letter combinations that the number could represent.
A mapping of digit to letters (just like on the telephone buttons) is given below.

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.

code

public class Solution {    public ArrayList<String> letterCombinations(String digits) {        ArrayList<String> result = new ArrayList<String>();        if (digits == null || digits.equals("")) {            return result;        }        Map<Character, char[]> map = new HashMap<Character, char[]>();        map.put('0', new char[] {});        map.put('1', new char[] {});        map.put('2', new char[] { 'a', 'b', 'c' });        map.put('3', new char[] { 'd', 'e', 'f' });        map.put('4', new char[] { 'g', 'h', 'i' });        map.put('5', new char[] { 'j', 'k', 'l' });        map.put('6', new char[] { 'm', 'n', 'o' });        map.put('7', new char[] { 'p', 'q', 'r', 's' });        map.put('8', new char[] { 't', 'u', 'v'});        map.put('9', new char[] { 'w', 'x', 'y', 'z' });        StringBuilder sb = new StringBuilder();        helper(map, digits, sb, result);        return result;    }    private void helper(Map<Character, char[]> map, String digits,         StringBuilder sb, ArrayList<String> result) {        if (sb.length() == digits.length()) {            result.add(sb.toString());            return;        }        for (char c : map.get(digits.charAt(sb.length()))) {            sb.append(c);            helper(map, digits, sb, result);            sb.deleteCharAt(sb.length() - 1);        }    }}