面试题目:手写一个LRU算法实现

@Slf4jpublicclass LRUCache<K, V>extends LinkedHashMap<K, V> {privateint cacheSize;public LRUCache(int cacheSize) {super(16, 0.75f,true);this.cacheSize = cacheSize;     }      @Overridepublicsynchronized V get(Object key) {returnsuper.get(key);     }      @Overridepublicsynchronized V put(K key, V value) {returnsuper.put(key, value);     }      @Overrideprotectedboolean removeEldestEntry(Map.Entry<K, V> eldest) {boolean f = size() > cacheSize;if (f) {             log.info("LRUCache清除第三方密钥缓存Key:[{}]", eldest.getKey());         }return f;     }publicstaticvoid main(String[] args) {         LRUCache<String, Object> cache =new LRUCache<>(5);         cache.put("A","A");         cache.put("B","B");         cache.put("C","C");         cache.put("D","D");         cache.put("E","E");         System.out.println("初始化:" + cache.keySet());         System.out.println("访问值:" + cache.get("C"));         System.out.println("访问C后:" + cache.keySet());         System.out.println("PUT F后:" + cache.put("F","F"));         System.out.println(cache.keySet());     }  }

//使用父类中的构造，初始化容量和加载因子，该初始化容量是指数组大小。public LinkedHashMap(int initialCapacity,float loadFactor) {super(initialCapacity, loadFactor);         accessOrder=false;     }//一个参数的构造public LinkedHashMap(int initialCapacity) {super(initialCapacity);         accessOrder=false;     }//无参构造public LinkedHashMap() {super();         accessOrder=false;     }//这个不用多说，用来接受map类型的值转换为LinkedHashMappublic LinkedHashMap(Map<?extends K, ?extends V> m) {super(m);         accessOrder=false;     }//真正有点特殊的就是这个，多了一个参数accessOrder。存储顺序，LinkedHashMap关键的参数之一就在这个，//true：指定迭代的顺序是按照访问顺序(近期访问最少到近期访问最多的元素)来迭代的。 false：指定迭代的顺序是按照插入顺序迭代，也就是通过插入元素的顺序来迭代所有元素//如果你想指定访问顺序，那么就只能使用该构造方法，其他三个构造方法默认使用插入顺序。public LinkedHashMap(int initialCapacity,float loadFactor,boolean accessOrder) {super(initialCapacity, loadFactor);this.accessOrder = accessOrder;     }

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public LinkedHashMap(int initialCapacity,float loadFactor,boolean accessOrder) {super(initialCapacity, loadFactor);this.accessOrder = accessOrder;     }