Problem Link:

http://oj.leetcode.com/problems/lru-cache/

Long long ago, I had a post for implementing the LRU Cache in C++ in my homepage:

http://www.cs.uml.edu/~jlu1/doc/codes/lruCache.html

So this time, I am trying to use Python to implement a LRU Cache.

However, the result is Time Limit Exceeded.....

My previous post is of C++ template version, but for this problem, the key and value are both integer, so I had a simpler version in C++ here.

I will keep trying to pass the test using python... damn efficiency...

#include <iostream>
#include <vector>
#include <map> using namespace std; struct LRUCacheEntry
{
int key;
int data;
LRUCacheEntry* prev;
LRUCacheEntry* next;
}; class LRUCache{
private:
map<int, LRUCacheEntry*> _mapping;
vector<LRUCacheEntry*> _freeEntries;
LRUCacheEntry * head;
LRUCacheEntry * tail;
LRUCacheEntry * entries; public:
LRUCache(int capacity) {
entries = new LRUCacheEntry[capacity];
for (int i=0; i<capacity; i++)
_freeEntries.push_back(entries+i);
head = new LRUCacheEntry;
tail = new LRUCacheEntry;
head->prev = NULL;
head->next = tail;
tail->next = NULL;
tail->prev = head;
} ~LRUCache()
{
delete head;
delete tail;
delete [] entries;
} int get(int key) {
LRUCacheEntry* node = _mapping[key];
if(node)
{
detach(node);
attach(node);
return node->data;
}
else return -1;
} void set(int key, int value) {
LRUCacheEntry* node = _mapping[key];
if(node)
{
// Move the node to the head and update the value
detach(node);
node->data = value;
attach(node);
}
else{
if ( _freeEntries.empty() )
{
// Get the last node
node = tail->prev;
// Move it to the head and update (key, value)
// Update the map
detach(node);
_mapping.erase(node->key);
node->data = value;
node->key = key;
_mapping[key] = node;
attach(node);
}
else{
node = _freeEntries.back();
_freeEntries.pop_back();
node->key = key;
node->data = value;
_mapping[key] = node;
attach(node);
}
}
} private:
void detach(LRUCacheEntry* node)
{
node->prev->next = node->next;
node->next->prev = node->prev;
}
void attach(LRUCacheEntry* node)
{
node->next = head->next;
node->prev = head;
head->next = node;
node->next->prev = node;
}
};

【LeetCode OJ】LRU Cache的更多相关文章

  1. 【LeetCode OJ】Interleaving String

    Problem Link: http://oj.leetcode.com/problems/interleaving-string/ Given s1, s2, s3, find whether s3 ...

  2. 【LeetCode OJ】Reverse Words in a String

    Problem link: http://oj.leetcode.com/problems/reverse-words-in-a-string/ Given an input string, reve ...

  3. LeetCode OJ:LRU Cache(最近使用缓存)

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

  4. 【LeetCode OJ】Validate Binary Search Tree

    Problem Link: https://oj.leetcode.com/problems/validate-binary-search-tree/ We inorder-traverse the ...

  5. 【LeetCode OJ】Recover Binary Search Tree

    Problem Link: https://oj.leetcode.com/problems/recover-binary-search-tree/ We know that the inorder ...

  6. 【LeetCode OJ】Same Tree

    Problem Link: https://oj.leetcode.com/problems/same-tree/ The following recursive version is accepte ...

  7. 【LeetCode OJ】Symmetric Tree

    Problem Link: https://oj.leetcode.com/problems/symmetric-tree/ To solve the problem, we can traverse ...

  8. 【LeetCode OJ】Binary Tree Level Order Traversal

    Problem Link: https://oj.leetcode.com/problems/binary-tree-level-order-traversal/ Traverse the tree ...

  9. 【LeetCode OJ】Binary Tree Zigzag Level Order Traversal

    Problem Link: https://oj.leetcode.com/problems/binary-tree-zigzag-level-order-traversal/ Just BFS fr ...

随机推荐

  1. hdu---(4515)小Q系列故事——世界上最遥远的距离(模拟题)

    小Q系列故事——世界上最遥远的距离 Time Limit: 500/200 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)T ...

  2. Tarjan--LCA算法的个人理解即模板

    tarjan---LCA算法的步骤是(当dfs到节点u时): 实际:  并查集+dfs 具体步骤: 1 在并查集中建立仅有u的集合,设置该集合的祖先为u 1 对u的每个孩子v:    1.1 tarj ...

  3. 如何在Objective-C中实现链式语法

    在接触到开源项目 Masonry 后,里面的布局约束的链式写法让我颇感兴趣,就像下面这样: 1 2 3 4 5 6 7 8 UIEdgeInsets padding = UIEdgeInsetsMak ...

  4. SVMtoy

    SVMtoy [label_matrix, instance_matrix] = libsvmread('ex8b.txt'); options = ''; % contour_level = [-1 ...

  5. Subscribe的第四个参数用法

    看别人的代码真的是很好的学习过程啊 之前用Subscribe订阅的时候都是简单的用法形如: ros::Subscriber sub = node.subscribe<uhf_rfid_api:: ...

  6. List<object> isEmpy contail 的判断

  7. tomcat 创建虚拟主机

    1. tomcat8 2. TOMCATROOT/conf/server.xml 增加<Host name="HOSTNAME" appBase="ROOTDir& ...

  8. 修改linux 文件权限命令 chmod

    [转载自:http://www.cnblogs.com/avril/archive/2010/03/23/1692809.html] Linux系统中的每个文件和目录都有访问许可权限,用它来确定谁可以 ...

  9. qt--- vs

    qt with vs 1.安装vs2012: 2.下载Qt 5.2.0 for Windows 32-bit (VS 2012, 579 MB) 和 Visual Studio Add-in 1.2. ...

  10. Xcode连接git@osc

    Xcode 已经集成了git,建立新项目时钩选使用git,然后按照下面步骤让Xcode和git@osc 建立连接. 第一步:成生SSH密钥 打开终端命令工具,输入命令:ssh-keygen -t rs ...