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”].

Recursive solution

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public class Solution {
    private static final String[] KEYS = {"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
    public List<String> letterCombinations(String digits) {
        List<String> result = new LinkedList<String>();
        if (digits == null || digits.length() == 0) return result;
        combine(result, new StringBuilder(), digits, 0);
        return result;
    }
    
    public void combine(List<String> result, StringBuilder sb, String digits, int index) {
        if (sb.length() >= digits.length()) {
            result.add(sb.toString());
        } else {
            String letters = KEYS[digits.charAt(index) - '0'];
            for (int i = 0; i < letters.length(); i++) {
                sb.append(letters.charAt(i));
                combine(result, sb, digits, index + 1);
                sb.setLength(sb.length() - 1);
            }
        }
    }
}

Iterative solution

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public class Solution {
    public List<String> letterCombinations(String digits) {
        LinkedList<String> ans = new LinkedList<String>();
        if (digits == null || digits.length() == 0) {
            return ans;
        }
        String[] mapping = new String[] {"0", "1", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
        ans.add("");
        for(int i =0; i<digits.length();i++){
            int x = Character.getNumericValue(digits.charAt(i));
            while(ans.peek().length()==i){
                String t = ans.remove();
                for(char s : mapping[x].toCharArray())
                    ans.add(t+s);
            }
        }
        return ans;
    }
}