高频题，需要用unordered_map和list，做到O(1)

需要注意的一点是，list用了splice改变了节点的位置，但是iterator并不会失效，这也代表unordered_map不需要进行更新。（可以把iterator当成指针方便理解）

class LRUCache {public:    int cap;    // recent visit will be moved to the head    list
     
      
       > l; // pair(key,val)    unordered_map
       
        
         
          >::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); */