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
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
}