146. LRU Cache[hard]

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146. LRU Cache

Total Accepted: 86514
Total Submissions: 545459
Difficulty: Hard

Design and implement a data structure for Least Recently Used (LRU) cache. It should support the following operations:get and set.

get(key) - Get the value (will always be positive) of the key if the key exists in the cache, otherwise return -1.
set(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.

很简单的题目

用三个map分别装：

1、Cache里面的数据

2、每个数据最后的更新时间（used_time）

3、每个时间分别对哪个数据进行了操作（note）

对于每一个操作维护好这三个map就行

class LRUCache{private:    int Capacity;    int Clock = 0;    map <int , int > Cache;    map <int , int > used_time;    map <int , int > note;public:            LRUCache(int capacity)     {        Capacity = capacity;        Clock = 0;    }        void note_check()    {        while( Cache.count( note.begin()->second ) == 0 ||                 used_time[ note.begin()->second ] != note.begin()->first    )            note.erase( note.begin() );    }        void check()    {        note_check();        while( Cache.size() > Capacity )        {            int x = note.begin()->second;            Cache.erase(x);            used_time.erase(x);            note.erase( note.begin() );            note_check();        }    }        int get(int key)     {        ++Clock;        if(Cache.count(key)!=0)        {            used_time[key] = Clock;            note[Clock] = key;            note_check();            return Cache[key];        }        return -1;    }        void set(int key, int value) {        ++Clock;        Cache[key] = value;        used_time[key] = Clock;        note[Clock] = key;        check();    }};

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