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[LeetCode] LFU Cache 最近最不常用页面置换缓存器
Design and implement a data structure for Least Frequently Used (LFU) 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 reaches its capacity, it should invalidate the least frequently used item before inserting a new item. For the purpose of this problem, when there is a tie (i.e., two or more keys that have the same frequency), the least recently used key would be evicted.
Follow up:
Could you do both operations in O(1) time complexity?
Example:
LFUCache cache = new LFUCache( 2 /* capacity */ );cache.put(1, 1);cache.put(2, 2);cache.get(1); // returns 1cache.put(3, 3); // evicts key 2cache.get(2); // returns -1 (not found)cache.get(3); // returns 3.cache.put(4, 4); // evicts key 1.cache.get(1); // returns -1 (not found)cache.get(3); // returns 3cache.get(4); // returns 4
s
[LeetCode] LFU Cache 最近最不常用页面置换缓存器