M146.LRU缓存

hash table, doubly-linked list, https://leetcode.cn/problems/lru-cache/

思路：先自己写一个链表，用字典存储键值和其对应的地址，然后完成函数即可，加上学习，用时约30min

struct Listnode
{
    int key, value;
    Listnode* next, * prev;
    Listnode(int k, int v) :key(k), value(v), next(NULL), prev(NULL) {}
    Listnode() :key(0), value(0), next(NULL), prev(NULL) {}
};

class LRUCache 
{
private:
    unordered_map<int, Listnode*> map;
    Listnode* head, * tail;
    int capacity, size;
public:
    
    LRUCache(int capacity)
    {
        this->capacity = capacity;
        this->size = 0;
        head = new Listnode();//伪头节点
        tail = new Listnode();
        head->next = tail;
        tail->prev = head;
    }

    int get(int key)
    {
        if (map.find(key) == map.end())
            return -1;
        Listnode* node = map[key];
        toHead(node);
        return node->value;
    }

    void put(int key, int value)
    {
        if (map.find(key) == map.end())//不存在
        {
            Listnode* node = new Listnode(key, value);
            map[key] = node;
            addNodeToHead(node);
            size++;
            if (size > capacity)//超出容量
            {
                map.erase(tail->prev->key);
                removeTail();
                size--;
            }
        }
        else//存在
        {
            Listnode* node = map[key];
            node->value = value;
            toHead(node);
        }
    }

    void removeNode(Listnode* node)
    {
        node->prev->next = node->next;
        node->next->prev = node->prev;
    }

    void addNodeToHead(Listnode* node)
    {
        node->prev = head;
        node->next = head->next;
        head->next->prev = node;
        head->next = node;
    }

    void toHead(Listnode* node)
    {
        removeNode(node);
        addNodeToHead(node);
    }

    void removeTail()
    {
        Listnode* node = tail->prev;
        removeNode(node);
    }
};