LRU Cache
描述
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.
分析
为了使查找、插入和删除都有较高的性能,这题的关键是要使用一个双向链表和一个HashMap,因为:
- HashMap保存每个节点的地址,可以基本保证在
O(1)
时间内查找节点 - 双向链表能后在
O(1)
时间内添加和删除节点,单链表则不行
具体实现细节:
- 越靠近链表头部,表示节点上次访问距离现在时间最短,尾部的节点表示最近访问最少
- 访问节点时,如果节点存在,把该节点交换到链表头部,同时更新hash表中该节点的地址
- 插入节点时,如果cache的size达到了上限capacity,则删除尾部节点,同时要在hash表中删除对应的项;新节点插入链表头部
代码
C++的std::list
就是个双向链表,且它有个 splice()
方法,O(1)
时间,非常好用。
// LRU Cache
// 时间复杂度O(logn),空间复杂度O(n)
class LRUCache{
private:
struct CacheNode {
int key;
int value;
CacheNode(int k, int v) :key(k), value(v){}
};
public:
LRUCache(int capacity) {
this->capacity = capacity;
}
int get(int key) {
if (cacheMap.find(key) == cacheMap.end()) return -1;
// 把当前访问的节点移到链表头部,并且更新map中该节点的地址
cacheList.splice(cacheList.begin(), cacheList, cacheMap[key]);
cacheMap[key] = cacheList.begin();
return cacheMap[key]->value;
}
void set(int key, int value) {
if (cacheMap.find(key) == cacheMap.end()) {
if (cacheList.size() == capacity) { //删除链表尾部节点(最少访问的节点)
cacheMap.erase(cacheList.back().key);
cacheList.pop_back();
}
// 插入新节点到链表头部, 并且在map中增加该节点
cacheList.push_front(CacheNode(key, value));
cacheMap[key] = cacheList.begin();
} else {
//更新节点的值,把当前访问的节点移到链表头部,并且更新map中该节点的地址
cacheMap[key]->value = value;
cacheList.splice(cacheList.begin(), cacheList, cacheMap[key]);
cacheMap[key] = cacheList.begin();
}
}
private:
list<CacheNode> cacheList; // doubly linked list
unordered_map<int, list<CacheNode>::iterator> cacheMap;
int capacity;
};