460. LFU Cache

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

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<pair<int, list<int>>>::iterator it_pair;
        list<int>::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<pair<int, list<int>>>Flist;//Frequency List
    unordered_map<int, info>cache;
    
    void touch(unordered_map<int, info>::iterator it,int key){
        auto it_key=it->second.it_key;
        auto it_pair=it->second.it_pair;
        int freq=it_pair->first+1;
        
        it_pair->second.erase(it_key);
        if(it_pair->second.size()==0){
            it_pair=Flist.erase(it_pair);
        }
        else 
            std::advance(it_pair,1);
        if(it_pair==Flist.end()||it_pair->first!=freq)
            it_pair=Flist.insert(it_pair,{freq,{key}});
        else 
            it_pair->second.emplace_back(key);
        it->second.it_pair=it_pair;
        it->second.it_key=std::prev(it_pair->second.end());//it_pair->second.back();
    }
};

/**
 * Your LFUCache object will be instantiated and called as such:
 * LFUCache obj = new LFUCache(capacity);
 * int param_1 = obj.get(key);
 * obj.put(key,value);
 */