Given a digit string, return all possible letter combinations that the number could represent.

A mapping of digit to letters (just like on the telephone buttons) is given below.

Input:Digit string "23"
Output: ["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"].

Note:
Although the above answer is in lexicographical order, your answer could be in any order you want.

題目大意：給定一個由數字構成的字串，根據如圖所示的手機按鍵中的 數字->字串 的關係，求出所有可能字母的組合。例如，給定數字字串 “23”，因為2在手機按鍵上對應字串 “abc”，3在手機按鍵上對應字串 “def”，所以輸出由9個字串組成，分別是 "ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"。

解題思路：採用遞迴的思想，每次傳遞的引數分別是：輸入的數字字串digits，記錄要返回字串的集合res，當前遍歷到digits的位置pos，當前得到的字串curCombination。每次根據pos取得digits對應位置的數字，然後根據這個數字去字典裡面查詢手機按鍵上對應的字串，對字串中的每個字元，新增到curCombination的末尾，並

解題程式碼：（3ms，beats 78.36%）

class Solution {
    String[] dict = new String[] { "", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz" };

	public void letterCombination(String digits, List<String> res, int pos, String curCombination) {
		if (pos >= digits.length()) {
			res.add(curCombination);
			return;
		}
		for (char c : dict[digits.charAt(pos) - '0'].toCharArray()) {
			letterCombination(digits, res, pos + 1, curCombination + c);
		}
	}

	public List<String> letterCombinations(String digits) {
		if (digits == null || digits.length() == 0) {
			return Collections.emptyList();
		}
		List<String> res = new ArrayList<>();
		letterCombination(digits, res, 0, "");
		return res;
	}
}