【Leetcode146】 LRU Cache

 //高效的get决定了需要有hash
 //每次操作都要将被操作的数的最近访问信息进行调整，涉及随机位置的变换，即从中间移动到队首，需要list在o(1)中移动节点
 class LRUCache {
 public:
     list<pair<int, int>> l_;
     unordered_map<int, list<pair<int, int>>::iterator> m_;
     int capacity_;
     LRUCache(int capacity) {
        capacity_ = capacity;
     }

     int get(int key) {
        auto k = m_.find(key);
        if(k == m_.end()) return -1;
        l_.splice(l_.begin(), l_, k->second);//物理地址并没有变，所以key对应的list中的地址并没有变，只是在逻辑上成了头了
        return k->second->second;
     }

     void put(int key, int value) {
        auto p = m_.find(key);
        if(p != m_.end()){
            p->second->second = value;
            l_.splice(l_.begin(), l_, p->second);
        }else{
            if(l_.size() >= capacity_){
                int key_remove = l_.back().first;
                l_.pop_back();
                m_.erase(key_remove);
            }
            l_.insert(l_.begin(), make_pair(key, value));
            m_.insert(make_pair(key, l_.begin()));
        }
        return;
     }
 };

/**
 * 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);
 */