Leetcode LRU Cache

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.

解题：

我以为get不会更新的，结果get也会更新，唉。
思路就是用链表搞，设头尾指针，插入时有两种情况：
（1）插入的元素不存在队列中
    1）如果未达到capacity则直接头插法插入
    2）若达到capacity则头插法插入，删除队尾元素
（2）插入的元素存在队列中
    更新一下放到头
查询时等同于插入的元素存在队列中的情况
使用了map将key值映射到一个class中，该class包含value值和队列元素的指针
队列元素包含前后指针和key值

class LRUCache{public:    LRUCache(int capacity) {        this->cap = capacity;        head = new Node(0);        tail = new Node(0);        head->next = tail;        tail->pre = head;        mp.clear();    }    int get(int key) { //get will update the queue and put the element in front of head        if(mp.find(key) == mp.end()) return -1;        Node *now = mp[key]->node;        Node *preNow = now->pre;        Node *nxtNow = now->next;        if(preNow == head) return mp[key]->val;        preNow->next = nxtNow;        nxtNow->pre = preNow;        now->next = head->next;        head->next->pre = now;        head->next = now;        now->pre = head;        return mp[key]->val;    }    void set(int key, int value) {        if(mp.find(key) == mp.end() || mp[key]->val == -1) { //first used key or doesn't exist in the queue            Node *now = new Node(key);            if(cap) { //doesn't reached its capacity                cap --;                Node *nxt = head->next;                now->next = nxt;                nxt->pre = now;                head->next = now;                now->pre = head;            }            else {                //insert in the head                head->next->pre = now;                now->next = head->next;                now->pre = head;                head->next = now;                //delete the tail                Node *pre = tail->pre;                Node *prepre = pre->pre;                prepre->next = tail;                tail->pre = prepre;                //mp[pre->key]->val = -1; //mark doesn't exist in the queue                mp.erase(pre->key);                free(pre);            }            Two *two = new Two(value, now);            //mp[key] = two;            mp.insert(pair<int, Two *>(key, two));        }        else { //exist in the queue            Node *now = mp[key]->node;            mp[key]->val = value;            Node *preNow = now->pre;            Node *nxtNow = now->next;            if(preNow == head) return ;            preNow->next = nxtNow;            nxtNow->pre = preNow;            now->next = head->next;            head->next->pre = now;            head->next = now;            now->pre = head;        }    }private:    class Node {    public:        Node *pre, *next;        int key;        Node(int x) {            pre = NULL, next = NULL;            key = x;        }    };    class Two {    public:        int val;        Node *node;        Two(int val, Node *node) {            this->val = val;            this->node = node;        }    };    map<int, Two *> mp;    int cap;    Node *head, *tail;};