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.
The functions get
and put
must each run in O(1)
average time complexity.
Example 1:
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 LRUCache lRUCache = new LRUCache(2); lRUCache.put(1, 1); // cache is {1=1} lRUCache.put(2, 2); // cache is {1=1, 2=2} lRUCache.get(1); // return 1 lRUCache.put(3, 3); // LRU key was 2, evicts key 2, cache is {1=1, 3=3} lRUCache.get(2); // returns -1 (not found) lRUCache.put(4, 4); // LRU key was 1, evicts key 1, cache is {4=4, 3=3} lRUCache.get(1); // return -1 (not found) lRUCache.get(3); // return 3 lRUCache.get(4); // return 4
Constraints:
1 <= capacity <= 3000
0 <= key <= 104
0 <= value <= 105
- At most 2
* 105
calls will be made toget
andput
.
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