CODE 113: 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.

static char[][] cdigits = { { ' ' }, {}, { 'a', 'b', 'c' },{ 'd', 'e', 'f' }, { 'g', 'h', 'i' }, { 'j', 'k', 'l' },{ 'm', 'n', 'o' }, { 'p', 'q', 'r', 's' }, { 't', 'u', 'v' },{ 'w', 'x', 'y', 'z' } };public ArrayList<String> letterCombinations(String digits) {// Start typing your Java solution below// DO NOT write main() functionif (null == digits || "".equals(digits)) {ArrayList<String> results = new ArrayList<String>();String str = "";results.add(str);return results;}return dfs(digits, 0);}ArrayList<String> dfs(String digits, int i) {if (i >= digits.length()) {return new ArrayList<String>();}int a = Character.digit(digits.charAt(i), 10);if (1 == a) {return dfs(digits, i + 1);}ArrayList<String> results = new ArrayList<String>();for (int j = 0; j < cdigits[a].length; j++) {char c = cdigits[a][j];ArrayList<String> tmps = dfs(digits, i + 1);if (tmps.isEmpty()) {results.add(new StringBuilder().append(c).toString());} else {for (String str : tmps) {str = c + str;results.add(new String(str));}}}return results;}