146. LRU Cache
Description
Design a data structure that follows the constraints of a Least Recently Used (LRU) cache.
Implement the LRUCache
class:
LRUCache(int capacity)
Initialize the LRU cache with positive sizecapacity
.int get(int key)
Return the value of thekey
if the key exists, otherwise return-1
.void put(int key, int value)
Update the value of thekey
if thekey
exists. Otherwise, add thekey-value
pair to the cache. If the number of keys exceeds thecapacity
from this operation, evict the least recently used key.
Constraints
1 <= capacity <= 3000
0 <= key <= 3000
0 <= value <= 104
At most
3 * 104
calls will be made toget
andput
.
Approach
Links
YouTube
Examples
Input:
["LRUCache", "put", "put", "get", "put", "get", "put", "get", "get", "get"]
[[2], [1, 1], [2, 2], [1], [3, 3], [2], [4, 4], [1], [3], [4]]
Output:
[null, null, null, 1, null, -1, null, -1, 3, 4]
Explanation:

Solutions
/**
* Time complexity : O(1) both for put and get since all operations with
* ordered dictionary. get/in/set/move_to_end/popitem
* (get/containsKey/put/remove) are done in a constant time.
* Space complexity : O(capacity) since the space is used only for an
* ordered dictionary with at most capacity + 1 elements.
*/
class LRUCache extends LinkedHashMap<Integer, Integer> {
private int capacity;
public LRUCache(int capacity) {
super(capacity, 0.75F, true);
this.capacity = capacity;
}
public int get(int key) {
return super.getOrDefault(key, -1);
}
public void put(int key, int value) {
super.put(key, value);
}
@Override
protected boolean removeEldestEntry(Map.Entry<Integer, Integer> eldest) {
return size() > capacity;
}
}
/**
* 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);
*/
Follow up
Could you do
get
andput
inO(1)
time complexity?
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