最近做了一道面试题，想用javascript做出来，就是感觉差了那么一点，其各位大佬解答

Question

给定一个“Flatten” Dictionary 对象其键是点分割的，例如:{"A":1,"B.A":2,"B.B":3,"CC.D.E":4,"CC.D.F":5},实现一个函数，将其转换 为一个嵌套的字典对象，根据上面例子，嵌套对象结果为{A:1,"B":{"A":2,"B":3},"CC":{"D":{"E":4,"F":5}}}
 

Answer 1

    /**
     * {A：1, B.A:2 B.B:3 CC.D.E:4 CC.DF:5}
     * @param args
     */
    public static void main(String[] args) {
        Map<String, Integer> flatten = new HashMap<>();
        //生成例子数据
        flatten.put("A",1);
        flatten.put("B.A",2);
        flatten.put("B.B",3);
        flatten.put("CC.D.E",4);
        flatten.put("CC.D.F",5);

        Map<String, Object> newFlatten = new HashMap<>();
        flatten.forEach((key,val)->{
            recursion(key,newFlatten, val);
        });
        System.out.println(flatten);
        System.out.println(newFlatten);
    }

    /**
     * 递归方法
     * @param key
     * @param map
     * @param val
     */
    public  static void recursion(String key ,Map<String,Object> map,Integer val){
        boolean contains = key.contains(".");
        //最后一个赋值
        if(!contains){
            map.put(key,val);
        }else{
            String beforeKey = StrUtil.subBefore(key, ".", false);
            String afterKey = StrUtil.subAfter(key, ".", false);
            HashMap<String, Object> newMap = new HashMap<>();
            if (!map.containsKey(beforeKey)){
                map.put(beforeKey,newMap);
            }else {
                newMap = (HashMap<String, Object>) map.get(beforeKey);
            }
            recursion(afterKey,newMap,val);
        }
    }