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| 1 | +Design a data structure that follows the constraints of a Least Recently Used (LRU) cache. |
| 2 | + |
| 3 | +Implement the LRUCache class: |
| 4 | + |
| 5 | + |
| 6 | + LRUCache(int capacity) Initialize the LRU cache with positive size capacity. |
| 7 | + int get(int key) Return the value of the key if the key exists, otherwise return -1. |
| 8 | + 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. |
| 9 | + |
| 10 | + |
| 11 | +The functions get and put must each run in O(1) average time complexity. |
| 12 | + |
| 13 | + |
| 14 | +Example 1: |
| 15 | + |
| 16 | +Input |
| 17 | +["LRUCache", "put", "put", "get", "put", "get", "put", "get", "get", "get"] |
| 18 | +[[2], [1, 1], [2, 2], [1], [3, 3], [2], [4, 4], [1], [3], [4]] |
| 19 | +Output |
| 20 | +[null, null, null, 1, null, -1, null, -1, 3, 4] |
| 21 | + |
| 22 | +Explanation |
| 23 | +LRUCache lRUCache = new LRUCache(2); |
| 24 | +lRUCache.put(1, 1); // cache is {1=1} |
| 25 | +lRUCache.put(2, 2); // cache is {1=1, 2=2} |
| 26 | +lRUCache.get(1); // return 1 |
| 27 | +lRUCache.put(3, 3); // LRU key was 2, evicts key 2, cache is {1=1, 3=3} |
| 28 | +lRUCache.get(2); // returns -1 (not found) |
| 29 | +lRUCache.put(4, 4); // LRU key was 1, evicts key 1, cache is {4=4, 3=3} |
| 30 | +lRUCache.get(1); // return -1 (not found) |
| 31 | +lRUCache.get(3); // return 3 |
| 32 | +lRUCache.get(4); // return 4 |
| 33 | + |
| 34 | + |
| 35 | + |
| 36 | +Constraints: |
| 37 | + |
| 38 | + |
| 39 | + 1 <= capacity <= 3000 |
| 40 | + 0 <= key <= 104 |
| 41 | + 0 <= value <= 105 |
| 42 | + At most 2 * 105 calls will be made to get and put. |
| 43 | + |
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