如何设计一个简单的cache系统

如何设计一个简单的cache系统  （LRU）

2014-06-03 16:29:44|  分类： 默认分类|举报|字号 订阅

    





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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{
unordered_map<int, node*> mp;
int capacity;
int size;
node* head;
node* tail;
public:
LRUCache(int c){
if(c<0)return;
head = new node(-1,-1);
tail = new node(-1,-1);
head->next = tail;
tail->pre = head;
mp.clear();
capacity = c;
size = 0;
}
    
int get(int k) {
unordered_map<int, node*>::iterator it = mp.find(k);
if(it != mp.end()){
node* p = it->second;
p->pre->next = p->next;
p->next->pre = p->pre;
putToHead(p);
return p->value;
}
else
return -1;
}
    
void set(int k, int val) {
if(capacity < 1) return; 
unordered_map<int, node*>::iterator it = mp.find(k);
if(it != mp.end()){
node* p = it->second;
p->pre->next = p->next;
p->next->pre = p->pre;
putToHead(p);
p->value = val;
}else{
node* p = new node(k, val);
putToHead(p);
mp[k] = p;
size++;
if(size>capacity){
p = tail->pre;
tail->pre = p->pre;
p->pre->next = tail;
it = mp.find(p->key);
mp.erase(it);
delete p;
}
}
}

void putToHead(node* p)
{
p->next = head->next;
p->pre = head;
head->next->pre = p;
head->next = p;
}
};