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?
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
class LRUCache {
private:
list<pair<int, int>> lst;
unordered_map<int, list<pair<int, int>>::iterator> mp;
int cap;
void makeHot(list<pair<int, int>>::iterator it) {
pair<int, int> data = *it;
lst.erase(it);
lst.push_front(data);
mp[data.first] = lst.begin();
}
public:
LRUCache(int capacity) {
cap = capacity;
}
int get(int key) {
if (mp.find(key) == mp.end()) {
return -1;
}
int val = (*mp[key]).second;
makeHot(mp[key]);
return val;
}
void put(int key, int value) {
if (mp.find(key) != mp.end()) {
lst.erase(mp[key]);
} else if (lst.size() == cap && cap >= 1) {
int key = lst.back().first;
lst.pop_back();
mp.erase(key);
}
lst.push_front(make_pair(key, value));
mp[key] = lst.begin();
}
};
/**
* 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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