Design and implement a data structure for  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

解题思路：

I use two structure to solve this problem

list<pair<int, list<int>>> records the visit info(count & time) of each key, let's call it big-L

the most inner <int> is the key, and list<int> rank them according to visit-time, from old to new.

pair<int, list<int>> associate each list<int> with visit-count

e.g. {2, {a, b}} means both a & b has been visited for 2 times, but b has a newer last-visit-time

and the other structure unordered_map<int, info> keeps the value and iterators in big-L of each key

so everytime we have to remove one element, we remove the first <int> of first pair<int, list<int>> in big-L

and everytime we have to insert one element, just check if the first pair's count equals 1. If it is, use that pair, push new element to its back; otherwise, insert a new pair.

and everytime an element is visited, we move it from the old pair to the next one in big-L. if no following pair exists, or following pair's count is not old-count+1, just insert a new pair with old-count+1. don't forget to update the iterators in map :D

横向一个频率（从小到大）链表，纵向一个LRU(从旧到新),

class LFUCache {public:    struct info{        int val;        list
     
      

>>::iterator it_pair; list

::iterator it_key; }; LFUCache(int capacity) { _capacity=capacity; } int get(int key) { auto it = cache.find(key); if(it == cache.end())return -1; touch(it,key); return it->second.val; } void put(int key, int value) { auto it = cache.find(key); if(it != cache.end()){ touch(it,key); it->second.val=value; } else { if(_capacity==0)return ; if(cache.size() == _capacity){ auto it=Flist.front().second.begin(); cache.erase(*it); Flist.front().second.erase(it); if(Flist.front().second.size()==0) Flist.pop_front(); } if(Flist.empty()||Flist.front().first!=1) Flist.push_front({

1,{key}}); else Flist.front().second.emplace_back(key); cache[key]={value, Flist.begin(),std::prev(Flist.front().second.end())}; } }private: int _capacity; list