LRU Cache

Hard~30 min

Design a data structure that follows the constraints of a Least Recently Used (LRU) cache.

Implement the LRUCache class:

  • LRUCache(capacity) — Initialize the LRU cache with positive size capacity.
  • get(key) — Return the value of the key if it exists, otherwise return -1. This also marks the key as recently used.
  • put(key, value) — Update the value of the key if it exists. Otherwise, add the key-value pair to the cache. If the number of keys exceeds the capacity, evict the least recently used key.

The functions get and put must each run in O(1) average time complexity.

Examples

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 cache = new LRUCache(2); cache.put(1, 1); // cache is {1=1} cache.put(2, 2); // cache is {1=1, 2=2} cache.get(1); // return 1, cache is {2=2, 1=1} cache.put(3, 3); // evicts key 2, cache is {1=1, 3=3} cache.get(2); // return -1 (not found) cache.put(4, 4); // evicts key 1, cache is {3=3, 4=4} cache.get(1); // return -1 (not found) cache.get(3); // return 3 cache.get(4); // return 4

Constraints

  • 1 <= capacity <= 3000
  • 0 <= key <= 10^4
  • 0 <= value <= 10^5
  • At most 2 * 10^5 calls will be made to get and put
  • Expected time complexity: O(1) per operation
Code
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Output

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