高頻題，需要用unordered_map和list，做到O(1)

需要注意的一點是，list用了splice改變了節點的位置，但是iterator並不會失效，這也代表unordered_map不需要進行更新。（可以把iterator當成指標方便理解）

class LRUCache {
public:
    int cap;
    // recent visit will be moved to the head
    list<pair<int,int>> l; // pair(key,val)
    unordered_map<int,list<pair<int
 
,int>>::iterator> m; // key -> iterator in list
    
    LRUCache(int capacity) {
        cap = capacity;
    }
    
    int get(int key) {
        if (!m.count(key)) return -1;
        int res=(*m[key]).second;
        l.splice(l.begin(), l, m[key]); //move key to the first
        return
 
 res;
    }
    
    void put(int key, int value) {
        if (m.count(key)) l.erase(m[key]);
        
        l.push_front(make_pair(key,value));
        m[key] = l.begin();
        
        if (l.size()>cap){     
            m.erase(l.back().first);
            l.pop_back();
        }  
    }
};

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