LeetCode-146.LRU Cache

Design and implement a data structure for Least Recently Used (LRU) cache. It should support the following operations: get and put.

get(key) - Get the value (will always be positive) of the key if the key exists in the cache, otherwise return -1.
put(key, value) - Set or insert the value if the key is not already present. When the cache reached its capacity, it should invalidate the least recently used item before inserting a new item.

Follow up:
Could you do both operations in O(1) time complexity?

Example:

LRUCache cache = new LRUCache( 2 /* capacity */ );
cache.put(1, 1);
cache.put(2, 2);
cache.get(1);       // returns 1
cache.put(3, 3);    // evicts key 2
cache.get(2);       // returns -1 (not found)
cache.put(4, 4);    // evicts key 1
cache.get(1);       // returns -1 (not found)
cache.get(3);       // returns 3
cache.get(4);       // returns 4

使用Java的LinkedHashMap实现

class LRUCache {//mytip
    LinkedHashMap<Integer,Integer> lru;
    int cap = 0;
    
    public LRUCache(int capacity) {
        //构造函数 accessOrder = true表示按照访问次序，为false表示按照插入次序
        lru=new LinkedHashMap<Integer,Integer>(0,0.75f,true){

            @Override
            protected boolean removeEldestEntry(Map.Entry<Integer, Integer> eldest) {
                return size()>cap;
            }

        };
        cap = capacity;
    }
    
    public int get(int key) {
        Integer re = lru.get(key);
        return null==re?-1:re;
    }
    
    public void put(int key, int value) {
        lru.put(key,value);
    }
}