# [面试题 16.25. LRU 缓存](https://leetcode.cn/problems/lru-cache-lcci) [English Version](/lcci/16.25.LRU%20Cache/README_EN.md) ## 题目描述 <!-- 这里写题目描述 --> <p>设计和构建一个“最近最少使用”缓存,该缓存会删除最近最少使用的项目。缓存应该从键映射到值(允许你插入和检索特定键对应的值),并在初始化时指定最大容量。当缓存被填满时,它应该删除最近最少使用的项目。</p> <p>它应该支持以下操作: 获取数据 <code>get</code> 和 写入数据 <code>put</code> 。</p> <p>获取数据 <code>get(key)</code> - 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。<br> 写入数据 <code>put(key, value)</code> - 如果密钥不存在,则写入其数据值。当缓存容量达到上限时,它应该在写入新数据之前删除最近最少使用的数据值,从而为新的数据值留出空间。</p> <p><strong>示例:</strong></p> <pre>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 </pre> ## 解法 ### 方法一 <!-- tabs:start --> ```python class Node: def __init__(self, key=0, value=0): self.key = key self.value = value self.prev = None self.next = None class LRUCache: def __init__(self, capacity: int): self.cache = {} self.head = Node() self.tail = Node() self.capacity = capacity self.size = 0 self.head.next = self.tail self.tail.prev = self.head def get(self, key: int) -> int: if key not in self.cache: return -1 node = self.cache[key] self.move_to_head(node) return node.value def put(self, key: int, value: int) -> None: if key in self.cache: node = self.cache[key] node.value = value self.move_to_head(node) else: node = Node(key, value) self.cache[key] = node self.add_to_head(node) self.size += 1 if self.size > self.capacity: node = self.remove_tail() self.cache.pop(node.key) self.size -= 1 def move_to_head(self, node): self.remove_node(node) self.add_to_head(node) def remove_node(self, node): node.prev.next = node.next node.next.prev = node.prev def add_to_head(self, node): node.next = self.head.next self.head.next.prev = node self.head.next = node node.prev = self.head def remove_tail(self): node = self.tail.prev self.remove_node(node) return node # Your LRUCache object will be instantiated and called as such: # obj = LRUCache(capacity) # param_1 = obj.get(key) # obj.put(key,value) ``` ```java class LRUCache { class Node { int key; int value; Node prev; Node next; Node() { } Node(int key, int value) { this.key = key; this.value = value; } } private Map<Integer, Node> cache; private Node head; private Node tail; private int capacity; private int size; public LRUCache(int capacity) { cache = new HashMap<>(); this.capacity = capacity; head = new Node(); tail = new Node(); head.next = tail; tail.prev = head; } public int get(int key) { if (!cache.containsKey(key)) { return -1; } Node node = cache.get(key); moveToHead(node); return node.value; } public void put(int key, int value) { if (cache.containsKey(key)) { Node node = cache.get(key); node.value = value; moveToHead(node); } else { Node node = new Node(key, value); cache.put(key, node); addToHead(node); ++size; if (size > capacity) { node = removeTail(); cache.remove(node.key); --size; } } } private void moveToHead(Node node) { removeNode(node); addToHead(node); } private void removeNode(Node node) { node.prev.next = node.next; node.next.prev = node.prev; } private void addToHead(Node node) { node.next = head.next; head.next.prev = node; head.next = node; node.prev = head; } private Node removeTail() { Node node = tail.prev; removeNode(node); return node; } } /** * Your LRUCache object will be instantiated and called as such: * LRUCache obj = new LRUCache(capacity); * int param_1 = obj.get(key); * obj.put(key,value); */ ``` <!-- tabs:end --> <!-- end -->