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 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.

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 to get and put.

# @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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