146. LRU Cache
Difficulty: Medium
Topics: Design
Similar Questions:
Problem:
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.
The cache is initialized with a positive capacity.
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
Solutions:
class LRUCache {
public:
LRUCache(int capacity) {
this->capacity = capacity;
}
int get(int key) {
if (keyToNode.count(key) == 0) {
return -1;
}
int val = keyToNode[key]->second;
cache.splice(cache.begin(), cache, keyToNode[key]);
return val;
}
void put(int key, int value) {
if (keyToNode.count(key) > 0) {
cache.splice(cache.begin(), cache, keyToNode[key]);
keyToNode[key]->second = value;
return;
}
cache.insert(cache.begin(), {key, value});
keyToNode[key] = cache.begin();
if (cache.size() > capacity) {
int key = cache.back().first;
cache.pop_back();
keyToNode.erase(key);
}
}
private:
int capacity;
list<pair<int, int>> cache;
unordered_map<int, list<pair<int, int>>::iterator> keyToNode;
};
/**
* 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);
*/