LRU 缓存机制 #

题目 #

中文

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) 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

English

Design and implement a data structure for Least Recently Used (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

分析 #

基础

进阶

高阶

题解 #

C++

Java

JavaScript

Kotlin

PHP

Python2

Python3

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)

Ruby

Rust

Scala

Swift