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`
- At most 2`* 105` calls will be made to `get` and `put`.