LRU缓存

题目: 146. LRU 缓存

最近最少使用(Least Recently Used), 忘差不多了, 复习一下. 顺带一提 python 的 functools 提供 lru_cache 方法, 作为修饰器使用很方便.

维护一个链表, 遍历找 key 是否存在. 存在的话移到头部, 更新/获取 value; 不存在的话插在头上, 然后如果满了删掉尾巴即可.

题目要求 Get/Put 方法平均时间复杂度在O(1)上, 需要一个哈希表根据 key 直接找到节点. 需要双向链表简化插入/删除操作. 额外再记录下元素数量判断是否已满.

写出来时间只超过 18%, 看了下其他人的答案, 发现被废弃的节点可以复用, 省去额外的创建时间, 超过 82%.

type Node struct {
    Key int
    Val int
    Prev *Node
    Next *Node
}

type LRUCache struct {
    Size int
    Cap int
    Elements *Node
    Mapping map[int]*Node
}

func Constructor(capacity int) LRUCache {
    root := &Node{}
    root.Prev = root
    root.Next = root
    return LRUCache{
        Size: 0,
        Cap: capacity,
        Elements: root,
        Mapping: make(map[int]*Node),
    }
}

func (this *LRUCache) Get(key int) int {
    if node, exist := this.Mapping[key]; exist {
        // move to head
        node.Prev.Next = node.Next
        node.Next.Prev = node.Prev
        head := this.Elements
        head.Next.Prev = node
        node.Next = head.Next
        head.Next = node
        node.Prev = head
        return node.Val
    }
    return -1
}

func (this *LRUCache) Put(key int, value int)  {
    head := this.Elements
    if node, exist := this.Mapping[key]; exist {
        // move to head
        node.Prev.Next = node.Next
        node.Next.Prev = node.Prev
        head.Next.Prev = node
        node.Next = head.Next
        head.Next = node
        node.Prev = head
        node.Val = value
        return
    }
    var node *Node
    if this.Size >= this.Cap {
        // remove tail
        t := head.Prev
        t.Prev.Next = t.Next
        t.Next.Prev = t.Prev
        delete(this.Mapping, t.Key)
        node = t
    } else {
        this.Size++
        node = &Node{}
    }
    node.Key = key
    node.Val = value
    head.Next.Prev = node
    node.Next = head.Next
    node.Prev = head
    head.Next = node
    this.Mapping[key] = node
}