此代码是否适用于LFU（最不常用）+ LRU（最近最少使用）实现？

Question

我想实现LFU（Least Freq Used），但是当缓存中所有项目的频率相同时，我希望以LRU（最近最少使用）的方式从缓存中删除该项目。

以下是代码：

class LFUCache {
    int capacity;
    int freq;int key,value;
    long timeStamp; //needed for LRU 

    Map<Integer, LFUCache> hm = new HashMap<>();
    Queue<LFUCache> queue = new PriorityQueue<>(new Comparator<LFUCache>(){
        public int compare(LFUCache c1, LFUCache c2){
            int freq1=c1.freq;
            int freq2=c2.freq;

            if(freq1<freq2){
                return -1;
            }else if(freq1>freq2){
                return 1;
            }else{

                return Long.compare(c1.timeStamp, c2.timeStamp);
            }
        }
    });
    public LFUCache(int capacity) {
        this.capacity = capacity;
    }
    public LFUCache(int key, int value){
        this.key=key;
        this.value=value;
    }

    public int get(int key) {
        if(hm.containsKey(key)){
            //cache hit
            LFUCache item = hm.get(key); //get the orignal item
            int data = item.value;
            queue.remove(item); //remove orignal item 

            item.freq++;
            long time = System.currentTimeMillis();
            item.timeStamp = time;

            queue.add(item);//add again
            hm.put(key,item);
            return data;
        }else{
            return -1;//item not found in cache

        }
    }

    public void put(int key, int value) {
        if(capacity==0)return;

          if(hm.containsKey(key)){
            //cache hit
            LFUCache item = hm.get(key);
            queue.remove(item);

            item.freq++;
            item.value=value;
            long time = System.currentTimeMillis();
            item.timeStamp = time;

            queue.add(item);
            hm.put(key,item);
            return;
        }
        if(capacity<=hm.size()){ //exceeded capacity so remove LFU item 
            //remove by using LRU and LFU 

            LFUCache item = queue.poll();
            hm.remove(item.key);

        }
            //creating and adding new item
            LFUCache cacheItem = new LFUCache(key,value);
            cacheItem.freq=1;
            long time = System.currentTimeMillis();
            cacheItem.timeStamp = time;

            queue.add(cacheItem);
            hm.put(key,cacheItem);

    }
}

我想尝试：LFUCache obj = new LFUCache(capacity); 和obj.put(key,value);

另外，我想以低于O（n）的复杂度实现get和put函数。