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 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 <= 10^4
0 <= value <= 10^5
At most 2 * 105 calls will be made to get and put.
Solution
dict + heap
Use a dict to store the key-value pair, and also store the timestamp of the calling of functions. If the capacity is exceeded, then pop from the heap, compare the timestamp in heap with timestamp in dict, if they are the same, pop from the dict.
The time complexity of put function is not strictly o ( 1 ) o(1) o(1) though.
dict + double-sided linked list
Use head
and tail
for the linked list!!
Code
dict + heap
python3
class LRUCache:
import heapq
def __init__(self, capacity: int):
self.key_value = {}
self.time_heap = []
self.timestamp = 0
self.capacity = capacity
def get(self, key: int) -> int:
if key not in self.key_value:
return -1
self.timestamp += 1
self.key_value[key][1] = self.timestamp
heapq.heappush(self.time_heap, (self.timestamp, key))
return self.key_value[key][0]
def put(self, key: int, value: int) -> None:
self.timestamp += 1
self.key_value[key] = [value, self.timestamp]
heapq.heappush(self.time_heap, (self.timestamp, key))
while len(self.key_value) > self.capacity:
timestamp, key = heapq.heappop(self.time_heap)
if self.key_value[key][1] == timestamp:
self.key_value.pop(key)
# Your LRUCache object will be instantiated and called as such:
# obj = LRUCache(capacity)
# param_1 = obj.get(key)
# obj.put(key,value)
Linked List
python3
class LinkedList:
def __init__(self, key: int, val: int):
self.key = key
self.val = val
self.prev = None
self.next = None
class LRUCache:
def __init__(self, capacity: int):
self.head = LinkedList(-1, -1)
self.tail = LinkedList(-1, -1)
self.head.next = self.tail
self.tail.prev = self.head
self.capacity = capacity
self.key_value = {}
def _remove(self, node) -> None:
prev = node.prev
pn = node.next
prev.next = pn
pn.prev = prev
def _add(self, node) -> None:
self.tail.prev.next = node
node.prev = self.tail.prev
self.tail.prev = node
node.next = self.tail
def get(self, key: int) -> int:
if key not in self.key_value:
return -1
node = self.key_value[key]
self._remove(node)
self._add(node)
return node.val
def put(self, key: int, value: int) -> None:
if key in self.key_value:
self._remove(self.key_value[key])
node = LinkedList(key, value)
self._add(node)
self.key_value[key] = node
while len(self.key_value) > self.capacity:
head_next = self.head.next
self._remove(head_next)
self.key_value.pop(head_next.key)
# Your LRUCache object will be instantiated and called as such:
# obj = LRUCache(capacity)
# param_1 = obj.get(key)
# obj.put(key,value)