146. LRU Cache
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.
Follow up: Could you do get
and put
in O(1)
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 <= 3000
0 <= value <= 104
- At most
3 * 104
calls will be made toget
andput
.
# @lc code=start
using LeetCode
struct LRUCache
cap::Int
data::OrderedDict{Int, Int}
LRUCache(cap::Int) = new(cap, OrderedDict{Int, Int}())
end
function Base.getindex(cache::LRUCache, key)
key ∉ keys(cache.data) && return -1
res = cache.data[key]
delete!(cache.data, key)
cache.data[key] = res
return res
end
function Base.setindex!(cache::LRUCache, val, key)
if key in keys(cache.data)
delete!(cache.data, key)
cache.data[key] = val
else
length(cache.data) == cache.cap && delete!(cache.data, first(cache.data|> keys))
cache.data[key] = val
end
end
# @lc code=end
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