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.

struct node {
node* pre;
int key;
int value;
node* next;
node(int k, int v):key(k),value(v),pre(NULL),next(NULL) {};
}; class LRUCache {
map<int, node*> mp;
node* head;
node* tail;
int size;
int capacity;
public:
LRUCache(int c) {
if (c < )return;
head = new node(, );
tail = new node(, );
head->next = tail;
tail->pre = head;
mp.clear();
size = ;
capacity = c;
} int get(int k) {
map<int, node*>::iterator it = mp.find(k);
if (it != mp.end()) {
node* cur = (*it).second;
cur->pre->next = cur->next;
cur->next->pre = cur->pre;
putToHead(cur);
return cur->value;
} else
return -;
} void set(int k, int val) {
if (capacity < )return;
map<int, node*>::iterator it = mp.find(k);
if (it != mp.end()) {//find
node* cur = (*it).second;
cur->pre->next = cur->next;
cur->next->pre = cur->pre;
cur->value = val;
putToHead(cur);
} else {//not find
node* tmp = new node(k,val);
putToHead(tmp);
mp[k] = tmp;
if (size < capacity) {//size < capacity
size++;
} else {//size >= capacity
node* deltmp = tail->pre;
tail->pre = deltmp->pre;
deltmp->pre->next = tail;
it = mp.find(deltmp->key);
mp.erase(it);
delete deltmp;
}
}
}
void putToHead(node* cur)
{
cur->next = head->next;
cur->pre = head;
cur->next->pre = cur;
head->next = cur;
} };

[LeetCode] LRU Cache [Forward]的更多相关文章

  1. [LeetCode] LRU Cache 最近最少使用页面置换缓存器

    Design and implement a data structure for Least Recently Used (LRU) cache. It should support the fol ...

  2. [LeetCode]LRU Cache有个问题,求大神解答【已解决】

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

  3. LeetCode:LRU Cache

    题目大意:设计一个用于LRU cache算法的数据结构. 题目链接.关于LRU的基本知识可参考here 分析:为了保持cache的性能,使查找,插入,删除都有较高的性能,我们使用双向链表(std::l ...

  4. LeetCode——LRU Cache

    Description: Design and implement a data structure for Least Recently Used (LRU) cache. It should su ...

  5. LeetCode: LRU Cache [146]

    [题目] Design and implement a data structure for Least Recently Used (LRU) cache. It should support th ...

  6. LeetCode – LRU Cache (Java)

    Problem Design and implement a data structure for Least Recently Used (LRU) cache. It should support ...

  7. Leetcode: LRU Cache 解题报告

    LRU Cache  Design and implement a data structure for Least Recently Used (LRU) cache. It should supp ...

  8. Leetcode:LRU Cache,LFU Cache

    在Leetcode上遇到了两个有趣的题目,分别是利用LRU和LFU算法实现两个缓存.缓存支持和字典一样的get和put操作,且要求两个操作的时间复杂度均为O(1). 首先说一下如何在O(1)时间复杂度 ...

  9. leetcode LRU Cache python

    class Node(object): def __init__(self,k,x): self.key=k self.val=x self.prev=None self.next=None clas ...

随机推荐

  1. JavaScript小技巧整理篇(非常全)

    能够为大家提供这些简短而实用的JavaScript技巧来提高大家编程能力,这对于我来说是件很开心的事.每天仅花上不到2分钟的时间中,你将可 以读遍JavaScript这门可怕的语言所呈现给我们的特性: ...

  2. 集训第五周 动态规划 K题 背包

    K - 背包 Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Statu ...

  3. 13-看图理解数据结构与算法系列(Trie树)

    Trie树 Trie树,是一种搜索树,也称字典树或单词查找树,此外也称前缀树,因为某节点的后代存在共同的前缀.它的key都为字符串,能做到高效查询和插入,时间复杂度为O(k),k为字符串长度,缺点是如 ...

  4. fmt:formatDate的输出格式详解

    <fmt:formatDate value="${isoDate}" type="both"/> 2004-5-31 23:59:59 <fm ...

  5. HTML Imports & polyfill

    组件化浏览器的兼容性问题汇总 框架依赖的 Web 标准技术 优先级高 HTML Imports HTML tempaltes ES6 to ES5 (搭建模块开发环境) polyfill https: ...

  6. android开发里跳过的坑——图片文件上传失败

    使用的apache的httpclient的jar包,做的http图片上传,上传时,服务器总返文件格式不对.后来发现,是由于在创建FileBody时,使用了默认的ContentType引起的.所以服务器 ...

  7. codeforces Gym 100971 A、B、C、F、G、K、L

    A题  直接把问号全部变为陆地如果所有陆地连通    那么再逐个把刚才变成陆地的问号变为水如果依旧连通有多种解 为什么我的代码跑不过去,和网上的题解思路一模一样!!?? #include<cst ...

  8. hdu - 2851 Lode Runner (最短路)

    http://acm.hdu.edu.cn/showproblem.php?pid=2851 首先有n层,每层的路径都有一个起点和终点和对应的危险值,如果某两层之间有交集,就能从这一层上到另外一层,不 ...

  9. Java DynamoDB 增加、删除、修改、查询

    准备jar包 <dependency> <groupId>com.amazonaws</groupId> <artifactId>aws-java-sd ...

  10. Why It is so hard to explain or show some thing

    Why it is hard to explain something or learn something? For example, when I first know the hadoop, I ...