LeetCode刷题日志-17.电话号码的字母组合

纯暴力解法，digits有多长，就循环多少次进行字母组合

class Solution {
    public List<String> letterCombinations(String digits) {
        List<String> reslut = new ArrayList<>();
        if(digits.equals(""))
        return reslut;

       Map<Integer, String> letter_hash = new HashMap<Integer, String>() {{
            put(2, "abc");
            put(3, "def");
            put(4, "ghi");
            put(5, "jkl");
            put(6, "mno");
            put(7, "pqrs");
            put(8, "tuv");
            put(9, "wxyz");
        }};
        if(digits.length() == 1){
            for(char c : letter_hash.get(Integer.parseInt(String.valueOf(digits.charAt(0)))).toCharArray()){
                reslut.add(String.valueOf(c));
            }
        }
        
        if(digits.length() == 2){
            for(char c : letter_hash.get(Integer.parseInt(String.valueOf(digits.charAt(0)))).toCharArray()){
                StringBuilder re  =new StringBuilder();
                for(char d : letter_hash.get(Integer.parseInt(String.valueOf(digits.charAt(1)))).toCharArray()){
                    re.append(String.valueOf(c));
                    re.append(String.valueOf(d));
                    reslut.add(re.toString());
                    re.delete(0,re.length());
                }
            }
        }

        if(digits.length() == 3){
            for(char c : letter_hash.get(Integer.parseInt(String.valueOf(digits.charAt(0)))).toCharArray()){
                StringBuilder re  =new StringBuilder();
                for(char d : letter_hash.get(Integer.parseInt(String.valueOf(digits.charAt(1)))).toCharArray()){
                    for(char e : letter_hash.get(Integer.parseInt(String.valueOf(digits.charAt(2)))).toCharArray()){
                        re.append(String.valueOf(c));
                        re.append(String.valueOf(d));
                        re.append(String.valueOf(e));
                        reslut.add(re.toString());
                        re.delete(0,re.length());
                    }
                }
            }
        }

        if(digits.length() == 4){
            for(char c : letter_hash.get(Integer.parseInt(String.valueOf(digits.charAt(0)))).toCharArray()){
                StringBuilder re  =new StringBuilder();
                for(char d : letter_hash.get(Integer.parseInt(String.valueOf(digits.charAt(1)))).toCharArray()){
                    for(char e : letter_hash.get(Integer.parseInt(String.valueOf(digits.charAt(2)))).toCharArray()){
                        for(char f : letter_hash.get(Integer.parseInt(String.valueOf(digits.charAt(3)))).toCharArray()){
                            re.append(String.valueOf(c));
                            re.append(String.valueOf(d));
                            re.append(String.valueOf(e));
                            re.append(String.valueOf(f));
                            reslut.add(re.toString());
                            re.delete(0,re.length());
                        }
                    }
                }
            }
        }
    return reslut;
    }
}