【LRU Cache】cpp

题目

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

get(key) - Get the value (will always be positive) of the key if the key exists in the cache, otherwise return -1.
set(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.

代码

class LRUCache{
private:
    struct CacheNode{
        int key;
        int value;
        CacheNode(int k, int v) : key(k), value(v) {}
    };
    std::list<CacheNode> cachelist;
    std::map<int, std::list<CacheNode>::iterator> cacheMap;
    int capacity;
public:
    LRUCache(int capacity) {
        this->capacity = capacity;
    }
    
    int get(int key) {
        if ( cacheMap.find(key) == cacheMap.end() ) return -1;
        cachelist.splice(cachelist.begin(), cachelist, cacheMap[key]);
        cacheMap[key] = cachelist.begin();
        return cacheMap[key]->value;
    }
    
    void set(int key, int value) {
        if ( cacheMap.find(key)==cacheMap.end() )
        {
            if ( cachelist.size()==capacity )
            {
                cacheMap.erase(cachelist.back().key);
                cachelist.pop_back();
            }
            cachelist.push_front(CacheNode(key,value));
            cacheMap[key] = cachelist.begin();
        }
        else
        {
            cacheMap[key]->value = value;
            cachelist.splice(cachelist.begin(), cachelist, cacheMap[key]);
            cacheMap[key] = cachelist.begin();
        }
    }
};

Tips

这个题目直接参考的网上solution。

记录几个当时的疑问:

1. 为什么要结合list和hashmap两种数据结构,只用Hashmap一种数据结构不行么?

因为,如果cache满了,hashmap是无法知道哪个元素是“最不可能被访问的”,但是用双链表(std::list)这种结构却可以轻松确定这个事情。

2. 为什么CacheNode中要有key这个成员?

如果要去除“最不可能被访问的元素”,我们知道这个元素的本身value在list的最后一个位置,但是我们怎么知道这个要被删除的元素对应的hasmap中的位置呢?因此,我们需要在CacheNode这个结构体中保存key和value。这样就可以通过list最后一个元素,知道要删除的hashmap中的位置。

=================================================

第二次过这道题,比一次稍微熟练一些,但是基本还是写不出来。参照着之前的思路,又写了两边,加深印象。

class LRUCache{
    private:
        struct CacheNode
        {
            int key;
            int value;
            CacheNode(int k, int v): key(k), value(v){}
        };
        int capacity;
        list<CacheNode> cacheList;
        unordered_map<int, list<CacheNode>::iterator> cacheMap;
    public:
        LRUCache(int capacity) {
            this->capacity = capacity;
        }
    
        int get(int key) {
            if ( cacheMap.find(key)==cacheMap.end() ) return -1;
            cacheList.splice(cacheList.begin(), cacheList, cacheMap[key]);
            cacheMap[key] = cacheList.begin();
            return cacheMap[key]->value;
        }
    
        void set(int key, int value) {
            if ( cacheMap.find(key)==cacheMap.end() ) 
            {
                if ( cacheList.size()==capacity )
                {
                    cacheMap.erase(cacheList.back().key);
                    cacheList.pop_back();
                }
                cacheList.push_front(CacheNode(key,value));
                cacheMap[key] = cacheList.begin();
            }
            else
            {
                cacheMap[key]->value = value;
                cacheList.splice(cacheList.begin(), cacheList, cacheMap[key]);
                cacheMap[key] = cacheList.begin();
            }
        }
};

 

posted on 2015-05-01 21:57  承续缘  阅读(233)  评论(0编辑  收藏  举报

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