給定一個僅包含數字2-9的字串，返回所有它能表示的字母組合。答案可以按 任意順序 返回。

給出數字到字母的對映如下（與電話按鍵相同）。注意 1 不對應任何字母。

示例 1：

輸入：digits = "23"
輸出：["ad","ae","af","bd","be","bf","cd","ce","cf"]
示例 2：

輸入：digits = ""
輸出：[]

解法一：回溯法

遍歷所有可能的結果，返回成功的結果

List<String> res = new ArrayList<>();
    HashMap<Character, String> hash = new
 
 HashMap<>();

    public List<String> letterCombinations(String digits) {
        if (digits.length() == 0)
            return new ArrayList<>();
        createHash();
        letter(digits, 0, new StringBuilder());
        return res;

    }

    void letter(String digits, int level, StringBuilder tempres) {

        
 
if (level == digits.length()) {
            res.add(tempres.toString());
            return;
        }
        String str = hash.get(digits.charAt(level));
        for (int i = 0; i < str.length(); i++) {
            tempres.append(str.charAt(i));
            letter(digits, level + 1, tempres);
            tempres.deleteCharAt(tempres.length() 
 
- 1);
        }
    }

    void createHash() {
        hash.put('2', "abc");
        hash.put('3', "def");
        hash.put('4', "ghi");
        hash.put('5', "jkl");
        hash.put('6', "mno");
        hash.put('7', "pqrs");
        hash.put('8', "tuv");
        hash.put('9', "wxyz");
    }

時間複雜度：O(3m×4n)

空間複雜度：O(m+n)

解法二：佇列

 HashMap<Character, String> hash = new HashMap<>();

    public List<String> letterCombinations(String digits) {
        if(digits.length()==0)
        return new ArrayList<>();
        createHash();
        Deque<String> queue = new LinkedList<>();
        queue.add("");
        for (int i = 0; i < digits.length(); i++) {
            int size = queue.size();

            for (int j = 0; j < size; j++) {
                String nowdigit = hash.get(digits.charAt(i));
                String curqueue = queue.poll();
                for (int k = 0; k < nowdigit.length(); k++)
                    queue.add(curqueue + nowdigit.charAt(k));
            }

        }
        List<String> res = new ArrayList<>();
        int len = queue.size();
        for (int i = 0; i < len; i++)
            res.add(queue.poll());
        return res;

    }

    void createHash() {
        hash.put('2', "abc");
        hash.put('3', "def");
        hash.put('4', "ghi");
        hash.put('5', "jkl");
        hash.put('6', "mno");
        hash.put('7', "pqrs");
        hash.put('8', "tuv");
        hash.put('9', "wxyz");
    }

總結：心裡要有層次遍歷的樹的結構