Monogram algorithm succinctly -leetcode17- phone number

Title Description

题目描述：
	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。
	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。

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

Problem-solving ideas

Call recursive loop, each acquiring a first input character string, the character [key] to traverse all characters corresponding to splicing.
As of recursive conditions: input string is cut completely, which is the need to press the End key press

The sample code

public class Test {
    //存储最终结果
    private static List<String> list = new ArrayList<>();
    //保存手机按键，数组的索引代表键位数，内容则为按键对应的数组
    private static final String[] KEY_STR = {"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
    public static void main(String[] args) {
        String numStr = "23";
       letterCombinations(numStr);
        System.out.println("结果为：" + list);
    }

    private static void letterCombinations(String numStr) {
        callbackMethod("", numStr);
    }

    private static void callbackMethod(String s, String numStr) {
        if ("".equals(numStr)){
            list.add(s);
        }else {
            //记录当前遍历到的数【按键】
            int currNum = Integer.parseInt(numStr.substring(0, 1));
            //该按键对应的所有字符
            String keyCurrStr = KEY_STR[currNum];
            char[] keyCurrChars = keyCurrStr.toCharArray();
            //遍历该按键的字符
            for (int i=0; i<keyCurrStr.length(); i++){
                callbackMethod(s + keyCurrChars[i], numStr.substring(1));
            }
        }
    }
}

operation result