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
LRU缓存利用了这样的一种思想。LRU是Least Recently Used 的缩写，翻译过来就是“最近最少使用”，也就是说，LRU缓存把最近最少使用的数据移除，让给最新读取的数据。而往往最常读取的，也是读取次数最多的，所以，利用LRU缓存，我们能够提高系统的performance.
见程序注释