回溯法（5）

原题：

/** * 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. */

答案：

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;    }}