自己没搞出来，只能借鉴别人实现，自己再实现了

 递归解法

 public class LetterCombinations {List<String> strings = new ArrayList<>();StringBuilder builder = new StringBuilder();public List<String> letterCombinations(String digits) {if(digits.isEmpty()) return strings;String[][] mapper = {{"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"}};letterCombinations(mapper, 0, digits);return strings;}public void letterCombinations(String[][] mapper, int inx, String digits) {//这里是返回每个组合的关键，也是递归跳出的关键if(inx == digits.length()) {strings.add(builder.toString());return;}int index = digits.charAt(inx) - 50;for (int i = 0; i < mapper[index].length; i++) {builder.append(mapper[index][i]);letterCombinations(mapper, inx + 1, digits);//每次返回一个组合后，要把上次存进去的删掉，不然会串到一块builder.deleteCharAt(builder.length() - 1);}}
}

 非递归，直接遍历，一种巧妙的方式

    public static List<String> letterCombinations(String digits) {List<String> result = new ArrayList<>();if(digits == null || digits.length() == 0) {return result;}String[] mapping = new String[] {"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};result.add("");for(int i = 0; i < digits.length(); i++) {List<String> temp = new ArrayList<>();String letters = mapping[digits.charAt(i) - '0'];for(int j = 0; j < letters.length(); j++) {//上一轮拼接结果和这一轮的字符拼接for(String s : result) {temp.add(s + letters.charAt(j));}}//将上一轮拼接结果返回给下一轮去使用result = temp;}return result;}