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| 1 | +#### 146. LRU 缓存 |
| 2 | + |
| 3 | +难度:中等 |
| 4 | + |
| 5 | +--- |
| 6 | + |
| 7 | +请你设计并实现一个满足 [LRU (最近最少使用) 缓存](https://baike.baidu.com/item/LRU) 约束的数据结构。 |
| 8 | + |
| 9 | +实现 `LRUCache` 类: |
| 10 | + |
| 11 | +* `LRUCache(int capacity)` 以 **正整数** 作为容量 `capacity` 初始化 LRU 缓存 |
| 12 | +* `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` 。 |
| 13 | +* `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。 |
| 14 | + |
| 15 | +函数 `get` 和 `put` 必须以 `O(1)` 的平均时间复杂度运行。 |
| 16 | + |
| 17 | + **示例:** |
| 18 | + |
| 19 | +``` |
| 20 | +输入 |
| 21 | +["LRUCache", "put", "put", "get", "put", "get", "put", "get", "get", "get"] |
| 22 | +[[2], [1, 1], [2, 2], [1], [3, 3], [2], [4, 4], [1], [3], [4]] |
| 23 | +输出 |
| 24 | +[null, null, null, 1, null, -1, null, -1, 3, 4] |
| 25 | + |
| 26 | +解释 |
| 27 | +LRUCache lRUCache = new LRUCache(2); |
| 28 | +lRUCache.put(1, 1); // 缓存是 {1=1} |
| 29 | +lRUCache.put(2, 2); // 缓存是 {1=1, 2=2} |
| 30 | +lRUCache.get(1); // 返回 1 |
| 31 | +lRUCache.put(3, 3); // 该操作会使得关键字 2 作废,缓存是 {1=1, 3=3} |
| 32 | +lRUCache.get(2); // 返回 -1 (未找到) |
| 33 | +lRUCache.put(4, 4); // 该操作会使得关键字 1 作废,缓存是 {4=4, 3=3} |
| 34 | +lRUCache.get(1); // 返回 -1 (未找到) |
| 35 | +lRUCache.get(3); // 返回 3 |
| 36 | +lRUCache.get(4); // 返回 4 |
| 37 | +``` |
| 38 | + |
| 39 | + **提示:** |
| 40 | + |
| 41 | +* `1 <= capacity <= 3000` |
| 42 | +* `0 <= key <= 10000` |
| 43 | +* `0 <= value <= 10^5` |
| 44 | +* 最多调用 `2 * 10^5` 次 `get` 和 `put` |
| 45 | + |
| 46 | +--- |
| 47 | + |
| 48 | +哈希表 + 双向链表: |
| 49 | + |
| 50 | + |
| 51 | + |
| 52 | +```Java |
| 53 | +type LRUCache struct { |
| 54 | + doubleLinkedList *list.List |
| 55 | + hashmap map[int]*list.Element |
| 56 | + capacity int |
| 57 | +} |
| 58 | + |
| 59 | +type entry struct { |
| 60 | + k, v int |
| 61 | +} |
| 62 | + |
| 63 | +func Constructor(capacity int) LRUCache { |
| 64 | + return LRUCache{list.New(), make(map[int]*list.Element), capacity} |
| 65 | +} |
| 66 | + |
| 67 | +func (this *LRUCache) Get(key int) int { |
| 68 | + node := this.hashmap[key] |
| 69 | + if node == nil { |
| 70 | + return -1 |
| 71 | + } |
| 72 | + this.doubleLinkedList.MoveToFront(node) |
| 73 | + return node.Value.(entry).v |
| 74 | +} |
| 75 | + |
| 76 | +func (this *LRUCache) Put(key int, value int) { |
| 77 | + node := this.hashmap[key] |
| 78 | + if node != nil { |
| 79 | + node.Value = entry{key, value} |
| 80 | + this.doubleLinkedList.MoveToFront(node) |
| 81 | + return |
| 82 | + } |
| 83 | + this.hashmap[key] = this.doubleLinkedList.PushFront(entry{key, value}) |
| 84 | + if this.capacity < len(this.hashmap) { |
| 85 | + delete(this.hashmap, this.doubleLinkedList.Remove(this.doubleLinkedList.Back()).(entry).k) |
| 86 | + } |
| 87 | +} |
| 88 | + |
| 89 | + |
| 90 | +/** |
| 91 | + * Your LRUCache object will be instantiated and called as such: |
| 92 | + * obj := Constructor(capacity); |
| 93 | + * param_1 := obj.Get(key); |
| 94 | + * obj.Put(key,value); |
| 95 | + */ |
| 96 | +``` |
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