# LRU 缓存机制
实现 LRUCache
类:
LRUCache(int capacity)
以正整数作为容量 capacity
初始化 LRU 缓存
int get(int key)
如果关键字 key
存在于缓存中,则返回关键字的值,否则返回 -1
。
void put(int key, int value)
如果关键字已经存在,则变更其数据值;如果关键字不存在,则插入该组「关键字-值」。当缓存容量达到上限时,它应该在写入新数据之前删除最久未使用的数据值,从而为新的数据值留出空间。
进阶:你是否可以在 O(1)
时间复杂度内完成这两种操作?
示例:
输入
["LRUCache", "put", "put", "get", "put", "get", "put", "get", "get", "get"]
[[2], [1, 1], [2, 2], [1], [3, 3], [2], [4, 4], [1], [3], [4]]
输出
[null, null, null, 1, null, -1, null, -1, 3, 4]
解释
LRUCache lRUCache = new LRUCache(2);
lRUCache.put(1, 1); // 缓存是 {1=1}
lRUCache.put(2, 2); // 缓存是 {1=1, 2=2}
lRUCache.get(1); // 返回 1
lRUCache.put(3, 3); // 该操作会使得关键字 2 作废,缓存是 {1=1, 3=3}
lRUCache.get(2); // 返回 -1 (未找到)
lRUCache.put(4, 4); // 该操作会使得关键字 1 作废,缓存是 {4=4, 3=3}
lRUCache.get(1); // 返回 -1 (未找到)
lRUCache.get(3); // 返回 3
lRUCache.get(4); // 返回 4
提示:
1 <= capacity <= 3000
0 <= key <= 10000
0 <= value <= 105
- 最多调用
2 * 105
次 get
和 put
以下错误的选项是?
## aop
### before
```c
#include
using namespace std;
```
### after
```c
```
## 答案
```c
class LRUCache
{
public:
LRUCache(int capacity) : cap(capacity) {}
int get(int key)
{
if (m.count(key) != 0)
{
int val = m[key]->second;
l.erase(m[key]);
l.push_front({key, val});
m[key] = l.begin();
return val;
}
return -1;
}
void put(int key, int value)
{
if (m.count(key) != 0)
{
l.erase(m[key]);
l.push_front({key, value});
m[key] = l.begin();
}
else
{
m.erase(l.back().first);
l.pop_back();
l.push_front({key, value});
m[key] = l.begin();
}
}
private:
int cap;
list> l;
unordered_map>::iterator> m;
};
```
## 选项
### A
```c
class LRUCache
{
private:
unordered_map cache;
DLinkedNode *head;
DLinkedNode *tail;
int size;
int capacity;
public:
LRUCache(int _capacity) : capacity(_capacity), size(0)
{
head = new DLinkedNode();
tail = new DLinkedNode();
head->next = tail;
tail->prev = head;
}
int get(int key)
{
if (!cache.count(key))
{
return -1;
}
DLinkedNode *node = cache[key];
moveToHead(node);
return node->value;
}
void put(int key, int value)
{
if (!cache.count(key))
{
DLinkedNode *node = new DLinkedNode(key, value);
cache[key] = node;
addToHead(node);
++size;
if (size > capacity)
{
DLinkedNode *removed = removeTail();
cache.erase(removed->key);
delete removed;
--size;
}
}
else
{
DLinkedNode *node = cache[key];
node->value = value;
moveToHead(node);
}
}
void addToHead(DLinkedNode *node)
{
node->prev = head;
node->next = head->next;
head->next->prev = node;
head->next = node;
}
void removeNode(DLinkedNode *node)
{
node->prev->next = node->next;
node->next->prev = node->prev;
}
void moveToHead(DLinkedNode *node)
{
removeNode(node);
addToHead(node);
}
DLinkedNode *removeTail()
{
DLinkedNode *node = tail->prev;
removeNode(node);
return node;
}
};
```
### B
```c
class LRUCache
{
private:
int capacity;
list> cl;
unordered_map>::iterator> cm;
public:
LRUCache(int capacity)
{
this->capacity = capacity;
}
int get(int key)
{
auto it = cm.find(key);
if (it != cm.end())
{
pair p = *cm[key];
int value = p.second;
cl.erase(cm[key]);
cl.push_front(p);
cm[key] = cl.begin();
return value;
}
else
return -1;
}
void put(int key, int value)
{
auto it = cm.find(key);
if (it != cm.end())
{
cl.erase(cm[key]);
cl.push_front({key, value});
cm[key] = cl.begin();
}
else
{
if (cl.size() == capacity)
{
int old_key = cl.back().first;
cl.pop_back();
cm.erase(old_key);
}
cl.push_front({key, value});
cm.insert({key, cl.begin()});
}
}
};
```
### C
```c
class LRUCache
{
private:
int capacity, size;
list cl;
unordered_map cm;
public:
LRUCache(int capacity)
{
this->capacity = capacity;
size = 0;
}
int get(int key)
{
auto it = cm.find(key);
if (it != cm.end())
{
cl.remove(key);
cl.push_back(key);
return it->second;
}
else
return -1;
}
void put(int key, int value)
{
auto it = cm.find(key);
if (it != cm.end())
{
it->second = value;
cl.remove(key);
cl.push_back(key);
}
else
{
if (size == capacity)
{
int old_key = cl.front();
cl.pop_front();
cm.erase(old_key);
size--;
}
cl.push_back(key);
cm.insert({key, value});
size++;
}
}
};
```