【LeetCode】LRU Cache

浏览数：31 / 时间：2015年06月20日

LRU Cache

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.

这题关键在于，怎样判断每个value是否算“最近使用”？

一个简单的想法是对每个键值对保留一个使用时间，当cache满时，删除最小时间对应的键值对。

问题在于寻找最小时间需要遍历，O(n)超时。

我们考虑使用空间换时间，建立一个双向链表，最近使用的调整到头部，需要删除则删除尾部。

代码实现如下：

struct BiListNode
{
    int key;
    BiListNode* prev;
    BiListNode* next;
    BiListNode(int k): key(k), prev(NULL), next(NULL){}
};

class LRUCache
{
public:
    int capacity;

    map<int, int> LRU;
    BiListNode* head;    //head point to the recently used (key,value)
    BiListNode* tail;    //tail point to the recently unused (key,value)
    map<int, BiListNode*> m;
    
    LRUCache(int capacity) 
    {
        this->capacity = capacity;
        head = NULL;
        tail = NULL;
    }
    
    int get(int key) 
    {
        map<int, int>::iterator it = LRU.find(key);
        if(it != LRU.end())
        {//find
            //update the sequence

            //the newly used node, move it to head
            BiListNode* target = (m.find(key))->second;
            if(target != head)
            {
                //target != head means target has prev
                target->prev->next = target->next;
                if(target->next != NULL)
                    target->next->prev = target->prev;
                else
                {//target is the tail
                    tail = target->prev;
                }

                target->next = head;
                head->prev = target;
                target->prev = NULL;

                //new head
                head = target;
            }

            return it->second;
        }
        else
            return -1;
    }
    
    void set(int key, int value) 
    {
        map<int, int>::iterator it = LRU.find(key);
        if(it != LRU.end())
        {//find, update
            //update the value
            it->second = value;
            
            //the newly used node, move it to head
            BiListNode* target = (m.find(key))->second;
            if(target != head)
            {
                //target != head means target has prev
                target->prev->next = target->next;
                if(target->next != NULL)
                    target->next->prev = target->prev;
                else
                {//target is the tail
                    tail = target->prev;
                }

                target->next = head;
                head->prev = target;
                target->prev = NULL;

                //new head
                head = target;
            }
        }
        else
        {//add, maybe delete first
            if(LRU.size() == capacity)
            {//full, delete first
                //delete from list
                if(tail->prev != NULL)
                    tail->prev->next = NULL;

                //delete from map
                map<int, BiListNode*>::iterator it = m.find(tail->key);
                m.erase(it);

                //delete from LRU
                map<int, int>::iterator it2 = LRU.find(tail->key);
                LRU.erase(it2);

                tail = tail->prev;
            }

            //add to list
            if(head == NULL)
            {
                head = new BiListNode(key);
                tail = head;
            }
            else
            {
                BiListNode* temp = new BiListNode(key);
                temp->next = head;
                head->prev = temp;
                head = temp;
            }

            //add to map
            m.insert(map<int, BiListNode*>::value_type(key, head));

            //add
            LRU.insert(map<int, int>::value_type(key, value));
            
        }
    }
};