17. 电话号码的字母组合

题目来源：17. 电话号码的字母组合 - 力扣（LeetCode）

描述：

给定一个仅包含数字 2-9 的字符串，返回所有它能表示的字母组合。答案可以按 任意顺序 返回。

给出数字到字母的映射如下（与电话按键相同）。注意 1 不对应任何字母。

示例 1：

输入：digits = "23"
输出：["ad","ae","af","bd","be","bf","cd","ce","cf"]

示例 2：

输入：digits = "2"
输出：["a","b","c"]

代码思路：

class Solution {
    public List<String> letterCombinations(String digits) {
        Map<Integer,String[]> map = new HashMap<>();
        List<String> res = new ArrayList<>();
        if(digits==null||digits.length()==0){
            return res;
        }
        for(int i = 0;i<digits.length();i++){
            Character ch = digits.charAt(i);
            switch (ch){
                case '2':map.put(i,new String[]{"a","b","c"});  break;
                case '3':map.put(i,new String[]{"d","e","f"});  break;
                case '4':map.put(i,new String[]{"g","h","i"});  break;
                case '5':map.put(i,new String[]{"j","k","l"});  break;
                case '6':map.put(i,new String[]{"m","n","o"});  break;
                case '7':map.put(i,new String[]{"p","q","r","s"});  break;
                case '8':map.put(i,new String[]{"t","u","v"});  break;
                case '9':map.put(i,new String[]{"w","x","y","z"});  break;
            }
        }


        backtring(0,digits.length(),map.get(0).length,res,map,"");
        return res;

    }
    public  void backtring(int index,int n,int k,List<String> res, Map<Integer,String[]> map,String st){
        if(index == n){
            res.add(new String(st));
            return;
        }
        for(int i=0;i<k;i++){
            st += map.get(index)[i];
            int h =map.getOrDefault(index+1,new String[]{}).length;
            backtring(index+1,n,h,res,map,st);
            st =  st.substring(0,st.length()-1);
        }

    }

}