Java LeetCode 17. 电话号码的字母组合

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。
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:
输入：“23”
输出：[“ad”, “ae”, “af”, “bd”, “be”, “bf”, “cd”, “ce”, “cf”].

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

       // map.put('4',"ghi"); 
       // map.put('5',"jkl"); 
       // map.put('6',"mno"); 
       // map.put('7',"pqrs"); 
      //  map.put('8',"tuv"); 
      //  map.put('9',"wxyz");
        
        StringBuilder str = new StringBuilder();
        back(0,map,digits,str,res);
        return res;
    }

    public void back(int start,Map<Character,char[]> map,String digits,StringBuilder str,List<String> res){
    
    
        if(digits.length()==start){
    
    
            res.add(str.toString());
            return;
        }
        char chars = digits.charAt(start);
        char[] s = map.get(chars);

        for(int i=0;i<s.length;i++){
    
    
            str.append(s[i]);

            back(start+1,map,digits,str,res);

            str.deleteCharAt(str.length()-1);
        }
    }   
}