77.LRU Cache（最近最久未使用算法）

Level：

  Hard

题目描述：

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

思路分析：

  设置一个map存放对应的键和值，同时设置一个双向链表，来保存最近最久未使用的关系，如果访问一个键，键存在于map中，访问完成后，我们在链表中将该键删除，然后将其添加到链表的首部，表示最近刚访问过这个键，当缓存满了后，如果要添加一个键值对，我们要删除的就是位于链表尾部的键和其对应的值，因为它是最久未访问的值。

代码：

class LRUCache {
    HashMap<Integer,Integer>cache=new HashMap<>();
    LinkedList<Integer>keys=new LinkedList<>();
    private static int sizeOfCache;
    public LRUCache(int capacity) {
        sizeOfCache=capacity;
    }

    public int get(int key) {
        if(cache.get(key)!=null){
            keys.remove(Integer.valueOf(key));//先在链表中删掉该键
            keys.addFirst(key); //然后将该键放到链表首部，表示刚被访问
            return cache.get(key);
        }
        return -1;
    }

    public void put(int key, int value) {
        if(cache.get(key)!=null)
            keys.remove(Integer.valueOf(key));
        else if(keys.size()==sizeOfCache){ //存储块已满
            int keyToRemove=keys.removeLast(); //链表最后一个键，代表最久未访问。
            cache.remove(keyToRemove);
        }
        keys.addFirst(key);
        cache.put(key,value);
    }
}

/**
 * Your LRUCache object will be instantiated and called as such:
 * LRUCache obj = new LRUCache(capacity);
 * int param_1 = obj.get(key);
 * obj.put(key,value);
 */