{ "data": { "question": { "questionId": "146", "questionFrontendId": "146", "categoryTitle": "Algorithms", "boundTopicId": 1074, "title": "LRU Cache", "titleSlug": "lru-cache", "content": "
Design a data structure that follows the constraints of a Least Recently Used (LRU) cache.
\n\nImplement 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.
\n
Example 1:
\n\n\nInput\n["LRUCache", "put", "put", "get", "put", "get", "put", "get", "get", "get"]\n[[2], [1, 1], [2, 2], [1], [3, 3], [2], [4, 4], [1], [3], [4]]\nOutput\n[null, null, null, 1, null, -1, null, -1, 3, 4]\n\nExplanation\nLRUCache lRUCache = new LRUCache(2);\nlRUCache.put(1, 1); // cache is {1=1}\nlRUCache.put(2, 2); // cache is {1=1, 2=2}\nlRUCache.get(1); // return 1\nlRUCache.put(3, 3); // LRU key was 2, evicts key 2, cache is {1=1, 3=3}\nlRUCache.get(2); // returns -1 (not found)\nlRUCache.put(4, 4); // LRU key was 1, evicts key 1, cache is {4=4, 3=3}\nlRUCache.get(1); // return -1 (not found)\nlRUCache.get(3); // return 3\nlRUCache.get(4); // return 4\n\n\n
\n
Constraints:
\n\n1 <= capacity <= 3000
0 <= key <= 104
0 <= value <= 105
2 * 105
calls will be made to get
and put
.LRUCache
类:LRUCache(int capacity)
以 正整数 作为容量 capacity
初始化 LRU 缓存int get(int key)
如果关键字 key
存在于缓存中,则返回关键字的值,否则返回 -1
。void put(int key, int value)
如果关键字 key
已经存在,则变更其数据值 value
;如果不存在,则向缓存中插入该组 key-value
。如果插入操作导致关键字数量超过 capacity
,则应该 逐出 最久未使用的关键字。函数 get
和 put
必须以 O(1)
的平均时间复杂度运行。
\n\n
示例:
\n\n\n输入\n[\"LRUCache\", \"put\", \"put\", \"get\", \"put\", \"get\", \"put\", \"get\", \"get\", \"get\"]\n[[2], [1, 1], [2, 2], [1], [3, 3], [2], [4, 4], [1], [3], [4]]\n输出\n[null, null, null, 1, null, -1, null, -1, 3, 4]\n\n解释\nLRUCache lRUCache = new LRUCache(2);\nlRUCache.put(1, 1); // 缓存是 {1=1}\nlRUCache.put(2, 2); // 缓存是 {1=1, 2=2}\nlRUCache.get(1); // 返回 1\nlRUCache.put(3, 3); // 该操作会使得关键字 2 作废,缓存是 {1=1, 3=3}\nlRUCache.get(2); // 返回 -1 (未找到)\nlRUCache.put(4, 4); // 该操作会使得关键字 1 作废,缓存是 {4=4, 3=3}\nlRUCache.get(1); // 返回 -1 (未找到)\nlRUCache.get(3); // 返回 3\nlRUCache.get(4); // 返回 4\n\n\n
\n\n
提示:
\n\n1 <= capacity <= 3000
0 <= key <= 10000
0 <= value <= 105
2 * 105
次 get
和 put
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out-of-bounds<\\/code>\\u548c
use-after-free<\\/code>\\u9519\\u8bef\\u3002<\\/p>\\r\\n\\r\\n
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Python 2.7.12<\\/code><\\/p>\\r\\n\\r\\n
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out-of-bounds<\\/code>\\u548c
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rust 1.58.1<\\/code><\\/p>\\r\\n\\r\\n
PHP 8.1<\\/code>.<\\/p>\\r\\n\\r\\n