leetcode 146. LRU Cache

Design and implement a data structure for Least Recently Used (LRU) cache. It should support the following operations: get and put.

get(key) - Get the value (will always be positive) of the key if the key exists in the cache, otherwise return -1.
put(key, value) - Set or insert the value if the key is not already present. When the cache reached its capacity, it should invalidate the least recently used item before inserting a new item.

The cache is initialized with a positive capacity.

Follow up:
Could you do both operations in O(1) time complexity?

Example:

LRUCache cache = new LRUCache( 2 /* capacity */ );

cache.put(1, 1);
cache.put(2, 2);
cache.get(1); // returns 1
cache.put(3, 3); // evicts key 2
cache.get(2); // returns -1 (not found)
cache.put(4, 4); // evicts key 1
cache.get(1); // returns -1 (not found)
cache.get(3); // returns 3
cache.get(4); // returns 4

• 思路：用链表存内容，然后用散列表存映射关系

class LRUCache {
public:
    int size;
    list<int> lru;
    unordered_map<int, list<int>::iterator> mp;
    unordered_map<int, int> kv;
    LRUCache(int capacity) {
        size=capacity;
    }
    void touch(int key){
        if(kv.count(key))
            lru.erase(mp[key]);
        lru.push_front(key);
        mp[key]=lru.begin();
    }
    int get(int key) {
        if(kv.count(key)==0)return -1;
        touch(key);
        return kv[key];
    }
    
    void put(int key, int value) {
        if(kv.size()==size&&kv.count(key)==0){// full
            mp.erase(lru.back());
            kv.erase(lru.back());
            lru.pop_back();
        }
        touch(key);
        kv[key]=value;
    }
};

/**
 * 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);
 */