{leetcode}/problems/lru-cache/[LeetCode - LRU Cache^]
使用双向链表实现时，使用两个"空节点" head 和 tail，可以有效减少空值判断，真是精巧。想起来上大学时，老师讲课提到这么一句。没想到竟然是如此实现。
LinkedHashMap 的实现也有不少的秘密可以探索。
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.
The cache is initialized with a positive capacity.
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
link:{sourcedir}/_0146_LRUCache.java[role=include]