leetcode146 - LRU Cache - medium

Design a data structure that follows the constraints of a Least Recently Used (LRU) cache.

Implement the LRUCache class:

• LRUCache(int capacity) Initialize the LRU cache with positive size capacity.
• int get(int key) Return the value of the key if the key exists, otherwise return -1.
• void put(int key, int value) Update the value of the key if the key exists. Otherwise, add the key-value pair to the cache. If the number of keys exceeds the capacity from this operation, evict the least recently used key.

Follow up:
Could you do get and put in O(1) time complexity?

 

Example 1:

Input
["LRUCache", "put", "put", "get", "put", "get", "put", "get", "get", "get"]
[[2], [1, 1], [2, 2], [1], [3, 3], [2], [4, 4], [1], [3], [4]]
Output
[null, null, null, 1, null, -1, null, -1, 3, 4]

Explanation
LRUCache lRUCache = new LRUCache(2);
lRUCache.put(1, 1); // cache is {1=1}
lRUCache.put(2, 2); // cache is {1=1, 2=2}
lRUCache.get(1);    // return 1
lRUCache.put(3, 3); // LRU key was 2, evicts key 2, cache is {1=1, 3=3}
lRUCache.get(2);    // returns -1 (not found)
lRUCache.put(4, 4); // LRU key was 1, evicts key 1, cache is {4=4, 3=3}
lRUCache.get(1);    // return -1 (not found)
lRUCache.get(3);    // return 3
lRUCache.get(4);    // return 4

 

Constraints:

• 1 <= capacity <= 3000
• 0 <= key <= 3000
• 0 <= value <= 104
• At most 3 * 104 calls will be made to get and put.

用DLL+Hashmap。

DLL track order, 新的在头旧的在尾，加到头删掉尾都是O(1).

Hashmap存key-val 查找O(1), 同时也存key在DLL里的位置，那given iterator删起来也是O(1).

Get/Put都要干的事：map里有key的话，删掉原位置。然后更新到队首。两个数据结构里都要更新都要更新！！

Put还要检查cache满没满，满了的话加条目之前先把队尾删了，map里对应的也删掉。

细节：

put的时候先看key存不存在，存在的话没必要删队尾的，还容易把它自己给删了，那你接下来找谁呢？所以这一步可以放到共同步骤里面

 

1. 作弊方法，用std的list，就是个DLL

class LRUCache {
public:
    unordered_map<int, pair<int, list<int>::iterator>> cache;
    list<int> order;
    int cap;
    LRUCache(int capacity) {
        cap=capacity;
    }
    
    void use(int key){
        if (cache.find(key) != cache.end()){
            order.erase(cache[key].second);
        }
        else if (order.size() >= cap){
            int oldest = order.back();
            order.pop_back();
            cache.erase(oldest);
        }
        order.push_front(key);
        cache[key].second = order.begin();
    }
    
    int get(int key) {
        if(cache.find(key) != cache.end()){
            use(key);
            return cache[key].first;
        }
        return -1;
    }
    
    void put(int key, int value) {
        use(key);
        cache[key].first = value;
    }
};