146. LRU Cache

题目

Design a data structure that follows the constraints of a Least Recently Used (LRU) cache.

Implement the LRUCache class:

LRUCache(int capacity) Initialize the LRU cache with positive size capacity.
int get(int key) Return the value of the key if the key exists, otherwise return 1.
void put(int key, int value) Update the value of the key if the key exists. Otherwise, add the key-value pair to the cache. If the number of keys exceeds the capacity from this operation, evict the least recently used key.

**Follow up:**Could you do get and put in O(1) time complexity?

Example 1:

Input
["LRUCache", "put", "put", "get", "put", "get", "put", "get", "get", "get"]
[[2], [1, 1], [2, 2], [1], [3, 3], [2], [4, 4], [1], [3], [4]]
Output
[null, null, null, 1, null, -1, null, -1, 3, 4]

Explanation
LRUCache lRUCache = new LRUCache(2);
lRUCache.put(1, 1); // cache is {1=1}
lRUCache.put(2, 2); // cache is {1=1, 2=2}
lRUCache.get(1);    // return 1
lRUCache.put(3, 3); // LRU key was 2, evicts key 2, cache is {1=1, 3=3}
lRUCache.get(2);    // returns -1 (not found)
lRUCache.put(4, 4); // LRU key was 1, evicts key 1, cache is {4=4, 3=3}
lRUCache.get(1);    // return -1 (not found)
lRUCache.get(3);    // return 3
lRUCache.get(4);    // return 4

Constraints:

1 <= capacity <= 3000
0 <= key <= 3000
0 <= value <= 104
At most 3 * 104 calls will be made to get and put.

题目大意

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。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) 如果关键字已经存在，则变更其数据值；如果关键字不存在，则插入该组「关键字-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？

解题思路

这一题是 LRU 经典面试题，详细解释见第三章 LRUCache 模板。

代码

package leetcode

type LRUCache struct {
	head, tail *Node
	Keys       map[int]*Node
	Cap        int
}

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

func Constructor(capacity int) LRUCache {
	return LRUCache{Keys: make(map[int]*Node), Cap: capacity}
}

func (this *LRUCache) Get(key int) int {
	if node, ok := this.Keys[key]; ok {
		this.Remove(node)
		this.Add(node)
		return node.Val
	}
	return -1
}

func (this *LRUCache) Put(key int, value int) {
	if node, ok := this.Keys[key]; ok {
		node.Val = value
		this.Remove(node)
		this.Add(node)
		return
	} else {
		node = &Node{Key: key, Val: value}
		this.Keys[key] = node
		this.Add(node)
	}
	if len(this.Keys) > this.Cap {
		delete(this.Keys, this.tail.Key)
		this.Remove(this.tail)
	}
}

func (this *LRUCache) Add(node *Node) {
	node.Prev = nil
	node.Next = this.head
	if this.head != nil {
		this.head.Prev = node
	}
	this.head = node
	if this.tail == nil {
		this.tail = node
		this.tail.Next = nil
	}
}

func (this *LRUCache) Remove(node *Node) {
	if node == this.head {
		this.head = node.Next
		node.Next = nil
		return
	}
	if node == this.tail {
		this.tail = node.Prev
		node.Prev.Next = nil
		node.Prev = nil
		return
	}
	node.Prev.Next = node.Next
	node.Next.Prev = node.Prev
}

Uh oh!

pFad - (p)hone/(F)rame/(a)nonymizer/(d)eclutterfier! Saves Data!