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  • [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

    code

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  • 原文地址:https://www.cnblogs.com/liuliu5151/p/10867184.html
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