146. LRU 缓存 ¶

题目¶

请你设计并实现一个满足 LRU (最近最少使用) 缓存约束的数据结构。

实现 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。

示例 1:

输入：

["LRUCache", "put", "put", "get", "put", "get", "put", "get", "get", "get"]

[[2], [1, 1], [2, 2], [1], [3, 3], [2], [4, 4], [1], [3], [4]]

输出：

[null, null, null, 1, null, -1, null, -1, 3, 4]

题解¶

双向链表

Go 解法 1Go 解法 2

type LRUCache struct {
    capacity int
    list     *list.List
    keyToNode map[int]*list.Element
}


type entry struct {
    key int
    val int
}


func Constructor(capacity int) LRUCache {
    return LRUCache{
        capacity: capacity,
        list: list.New(),
        keyToNode: map[int]*list.Element{},
    }
}


func (this *LRUCache) Get(key int) int {
    node := this.keyToNode[key]
    if node == nil {
        return -1
    }

    this.list.MoveToFront(node)
    return node.Value.(entry).val
}


func (this *LRUCache) Put(key int, value int)  {
    if node := this.keyToNode[key]; node != nil {
        // 更新
        node.Value = entry{key, value}
        this.list.MoveToFront(node)
        return
    }

    this.keyToNode[key] = this.list.PushFront(entry{key, value})
    if len(this.keyToNode) > this.capacity {
        delete(this.keyToNode, this.list.Remove(this.list.Back()).(entry).key)
    }
}

type node struct {
    key, val   int
    prev, next *node
}

type LRUCache struct {
    capacity   int
    cache      map[int]*node
    head, tail *node
}

func Constructor(capacity int) LRUCache {
    head := new(node)
    tail := new(node)
    head.next = tail
    tail.prev = head
    return LRUCache{
        capacity: capacity,
        cache:    make(map[int]*node, capacity),
        head:     head,
        tail:     tail,
    }
}

func (this *LRUCache) Get(key int) int {
    n, ok := this.cache[key]
    if !ok {
        return -1
    }
    this.moveToFront(n)
    return n.val
}

func (this *LRUCache) Put(key int, value int) {
    n, ok := this.cache[key]
    if ok {
        n.val = value
        this.moveToFront(n)
        return
    }
    if len(this.cache) == this.capacity {
        back := this.tail.prev
        this.remove(back)
        delete(this.cache, back.key)
    }
    n = &node{key: key, val: value}
    this.pushFront(n)
    this.cache[key] = n
}

func (this *LRUCache) moveToFront(n *node) {
    this.remove(n)
    this.pushFront(n)
}

func (this *LRUCache) remove(n *node) {
    n.prev.next = n.next
    n.next.prev = n.prev
    n.prev = nil
    n.next = nil
}

func (this *LRUCache) pushFront(n *node) {
    n.prev = this.head
    n.next = this.head.next
    this.head.next.prev = n
    this.head.next = n
}

Python 解法 1Python 解法 2

class Node:
    # 提高访问属性的速度，并节省内存
    __slots__ = 'prev', 'next', 'key', 'value'

    def __init__(self, key=0, value=0):
        self.key = key
        self.value = value

class LRUCache:
    def __init__(self, capacity: int):
        self.capacity = capacity
        self.dummy = Node()  # 哨兵节点
        self.dummy.prev = self.dummy
        self.dummy.next = self.dummy
        self.key_to_node = dict()

    def get_node(self, key: int) -> Optional[Node]:
        if key not in self.key_to_node:  # 没有这本书
            return None
        node = self.key_to_node[key]  # 有这本书
        self.remove(node)  # 把这本书抽出来
        self.push_front(node)  # 放在最上面
        return node

    def get(self, key: int) -> int:
        node = self.get_node(key)
        return node.value if node else -1

    def put(self, key: int, value: int) -> None:
        node = self.get_node(key)
        if node:  # 有这本书
            node.value = value  # 更新 value
            return
        self.key_to_node[key] = node = Node(key, value)  # 新书
        self.push_front(node)  # 放在最上面
        if len(self.key_to_node) > self.capacity:  # 书太多了
            back_node = self.dummy.prev
            del self.key_to_node[back_node.key]
            self.remove(back_node)  # 去掉最后一本书

    # 删除一个节点（抽出一本书）
    def remove(self, x: Node) -> None:
        x.prev.next = x.next
        x.next.prev = x.prev

    # 在链表头添加一个节点（把一本书放在最上面）
    def push_front(self, x: Node) -> None:
        x.prev = self.dummy
        x.next = self.dummy.next
        x.prev.next = x
        x.next.prev = x

class Node:
    def __init__(self, key=0, val=0):
        self.key = key
        self.val = val
        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.val

    def put(self, key: int, value: int) -> None:
        if key in self.cache:
            node = self.cache[key]
            node.val = 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
        node.prev = self.head
        self.head.next = node
        node.next.prev = node

    def remove_tail(self):
        node = self.tail.prev
        self.remove_node(node)
        return node

复杂度¶

时间复杂度： $O (1)$
空间复杂度： $O (m i n (p, c a p a c i t y))$ ，其中 $p$ 为 $p u t$ 的调用次数。

146. LRU 缓存¶

题目¶

题解¶

复杂度¶

参考¶

146. LRU 缓存 ¶