[leetcode]460. LFU Cache最低频率缓存
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 1 cache.put(3, 3); // evicts key 2 cache.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 3 cache.get(4); // returns 4
题意
与LRU不同，LFU的替换依据是其使用频率。Via wikipedia "When the cache is full and requires more room the system will purge the item with the lowest reference frequency."
Solution1
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