LRU 缓存机制 #

运用你所掌握的数据结构，设计和实现一个 [LRU (最近最少使用)](https://baike.baidu.com/item/LRU) 缓存机制。它应该支持以下操作： 获取数据 get 和 写入数据 put 。 获取数据 get(key) - 如果密钥 (key) 存在于缓存中，则获取密钥的值（总是正数），否则返回 -1。 写入数据 put(key, value) - 如果密钥已经存在，则变更其数据值；如果密钥不存在，则插入该组「密钥 / 数据值」。当缓存容量达到上限时，它应该在写入新数据之前删除最久未使用的数据值，从而为新的数据值留出空间。 进阶: 你是否可以在  `O(1)` 时间复杂度内完成这两种操作？ 示例: ``` LRUCache cache = new LRUCache( 2 /* 缓存容量 */ ); cache.put(1, 1); cache.put(2, 2); cache.get(1); // 返回 1 cache.put(3, 3); // 该操作会使得密钥 2 作废 cache.get(2); // 返回 -1 (未找到) cache.put(4, 4); // 该操作会使得密钥 1 作废 cache.get(1); // 返回 -1 (未找到) cache.get(3); // 返回 3 cache.get(4); // 返回 4 ```

Design and implement a data structure for [Least Recently Used (LRU)](https://en.wikipedia.org/wiki/Cache_replacement_policies#LRU) cache. It should support the following operations: get and put. get(key) - Get the value (will always be positive) of the key if the key exists in the cache, otherwise return -1. put(key, value) - Set or insert the value if the key is not already present. When the cache reached its capacity, it should invalidate the least recently used item before inserting a new item. The cache is initialized with a positive capacity. Follow up: Could you do both operations in O(1) time complexity? Example: ``` LRUCache cache = new LRUCache( 2 /* capacity */ ); cache.put(1, 1); cache.put(2, 2); cache.get(1); // returns 1 cache.put(3, 3); // evicts key 2 cache.get(2); // returns -1 (not found) cache.put(4, 4); // evicts key 1 cache.get(1); // returns -1 (not found) cache.get(3); // returns 3 cache.get(4); // returns 4 ```

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```python class DLinkedNode: def __init__(self): self.key = 0 self.val = 0 self.pre = None self.next = None class LRUCache: def _add(self, node: DLinkedNode): node.next = self.head.next node.pre = self.head self.head.next.pre = node self.head.next = node def _remove(self, node: DLinkedNode): if not node or not node.pre or not node.next: return node.pre.next = node.next node.next.pre = node.pre def _move_to_head(self, node: DLinkedNode): self._remove(node) self._add(node) def _pop_tail(self) -> DLinkedNode: if self.size == 0: return None node = self.tail.pre self._remove(node) return node def __init__(self, capacity: int): self.capacity = capacity self.size = 0 self.cache = {} self.head = DLinkedNode() self.tail = DLinkedNode() self.head.next = self.tail self.tail.pre = self.head def get(self, key: int) -> int: node = self.cache.get(key, None) if not node: return -1 self._move_to_head(node) return node.val def put(self, key: int, value: int) -> None: node = self.cache.get(key, None) if not node: node = DLinkedNode() node.key = key node.val = value self.cache[key] = node self.size = self.size + 1 self._add(node) if self.size > self.capacity: node = self._pop_tail() del(self.cache[node.key]) else: node.val = value self._move_to_head(node) ```

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