1. 题目
Design and implement a data structure for Least Recently Used (LRU) cache. It should support the following operations: get and put.
get(key) - Get the value (will always be positive) of the key if the key exists in the cache, otherwise return -1.
put(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.
Follow up:
Could you do both operations in O(1) time complexity?
Example:
LRUCache cache = new LRUCache( 2 / capacity / );
cache.put(1, 1);
cache.put(2, 2);
cache.get(1); // returns 1
cache.put(3, 3); // evicts key 2
cache.get(2); // returns -1 (not found)
cache.put(4, 4); // evicts key 1
cache.get(1); // returns -1 (not found)
cache.get(3); // returns 3
cache.get(4); // returns 4
2. 思路
因为要求是O(1)的复杂度,因此必须要用hash来索引。要求LRU规则的淘汰,因此建立一个链表,每次对key产生访问之后,就将此kv移动到最前端。容器不够时,从末端进行淘汰。
3. 代码
class LRUCache {
private:
typedef std::pair<int, int> Node;
typedef std::list<Node> Cont;
typedef Cont::iterator Iter;
typedef std::unordered_map<int, Iter> Idx;
Cont _cont;
Idx _idx;
int _cap;
int _num;
public:
LRUCache(int capacity) : _cap(capacity), _num(0) {
}
int get(int key) {
auto it = _idx.find(key);
if (it == _idx.end()) {
return -1;
}
Node n = *(it->second);
_cont.erase(it->second);
_cont.push_front(n);
_idx[key] = _cont.begin();
return n.second;
}
void put(int key, int value) {
auto it = _idx.find(key);
if (it != _idx.end()) {
_cont.erase(it->second);
_idx.erase(it);
_num--;
}
if (_num >= _cap) {
Node b = _cont.back();
_cont.pop_back();
_num -= _idx.erase(b.first);
}
Node n(key, value);
_cont.push_front(n);
_idx[key] = _cont.begin();
_num++;
return ;
}
};
/**
* Your LRUCache object will be instantiated and called as such:
* LRUCache obj = new LRUCache(capacity);
* int param_1 = obj.get(key);
* obj.put(key,value);
*/
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