【LeetCode】146. LRU Cache

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年龄，当cache满时，删除最“老”的键值对。

然而在删除节点时，寻找最“老”节点需要O(n)时间。

因此建立一个双向链表，最近使用的调整到头部，需要删除则删除尾部。

这样寻找最“老”节点就为O(1)时间。

然而在get函数时查找所需节点仍为O(n)时间。

因此再加入映射表m，空间换时间，查找变为O(1)。

struct Node
{
    int key;
    int val;
    Node* prev;
    Node* next;
    Node(int k, int v): key(k), val(v), prev(NULL), next(NULL) {}
};

class LRUCache{
public:
    Node* head; //most recently used
    Node* tail; //least recently used
    unordered_map<int, Node*> m;
    int curcap;
    int maxcap;

    LRUCache(int capacity) {
        head = new Node(-, -);
        tail = new Node(-, -);
        head->next = tail;
        tail->prev = head;
        curcap = ;
        maxcap = capacity;
    }

    int get(int key) {
        if(m[key] == NULL)
            return -;
        else
        {
            Node* node = m[key];
            delnode(node);
            addtohead(node);
            return node->val;
        }
    }

    void set(int key, int value) {
        if(m[key] == NULL)
        {
            if(curcap == maxcap)
            {
                m[tail->prev->key] = NULL;
                delnode(tail->prev);
            }
            Node* node = new Node(key, value);
            addtohead(node);
            m[key] = node;
            if(curcap < maxcap)
                curcap ++;
        }
        else
        {
            Node* node = m[key];
            node->val = value;
            delnode(node);
            addtohead(node);
        }
    }

    void delnode(Node* node)
    {
        Node* prev = node->prev;
        Node* next = node->next;
        prev->next = next;
        next->prev = prev;
    }

    void addtohead(Node* node)
    {
        node->next = head->next;
        node->prev = head;
        head->next->prev = node;
        head->next = node;
    }
};