LeetCode[One hard A Day] - LRU Cache

LRU Cache

题目需求：1）存入key and value, 可以通过key获得value

                        可以用hashmap来实现(unordered_map)

               2）当capacity不够时，remove掉least recently used

                        可以用double linked-list来实现(list)

class LRUCache {    
public:
    LRUCache(int capacity) {
        size = capacity;
    }
    
    int get(int key) {
        auto it = hash.find(key);
        if(it==hash.end()) return -1;
        cache.splice(cache.begin(), cache, it->second);
        return it->second->second;
    }
    
    void put(int key, int value) {
        auto it = hash.find(key);
        if(it!=hash.end()){
            it->second->second = value;
            return cache.splice(cache.begin(), cache, it->second);
        }
        cache.insert(cache.begin(), make_pair(key, value));
        hash[key] = cache.begin();
        if(cache.size()>size){
            hash.erase(cache.back().first);
            cache.pop_back();
        }
        
    }
private:
    unordered_map<int, list<pair<int, int>>::iterator> hash;
    list<pair<int, int>> cache;
    int size;
};

详细的大家可以看一下reference，感觉没有他讲的好～～～

reference: https://blog.youkuaiyun.com/qq508618087/article/details/50995188