0146. L R U Cache

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
}

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