Medium
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 size capacity
.int get(int key)
Return the value of the key
if the key exists, otherwise return -1
.void put(int key, int value)
Update the value of the key
if the key
exists. Otherwise, add the key-value
pair to the cache. If the number of keys exceeds the capacity
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
2 * 105
calls will be made to get
and put
.class LRUCache {
int length = 0;
Map cache = {};
LRUCache(int capacity) {
length = capacity;
}
int get(int key) {
if (cache[key] == null) {
return -1;
} else {
var value = cache[key];
cache.remove(key);
cache[key] = value;
return value;
}
}
void put(int key, int value) {
if (cache[key] != null) {
cache.remove(key);
}
cache[key] = value;
if (length < cache.keys.length) {
cache.remove(cache.keys.first);
}
}
}
/**
* Your LRUCache object will be instantiated and called as such:
* LRUCache obj = LRUCache(capacity);
* int param1 = obj.get(key);
* obj.put(key,value);
*/