题目
Design and implement a data structure for Least Frequently Used (LFU) cache.
Implement the LFUCache
class:
LFUCache(int capacity)
Initializes the object with the capacity
of the data structure.
int get(int key)
Gets the value of the key
if the key
exists in the cache. Otherwise, returns 1
.
void put(int key, int value)
Sets or inserts the value if the key
is not already present. When the cache reaches its capacity
, it should invalidate the least frequently used item before inserting a new item. For this problem, when there is a tie (i.e., two or more keys with the same frequency), the least recently used key
would be evicted.
Notice that the number of times an item is used is the number of calls to the get
and put
functions for that item since it was inserted. This number is set to zero when the item is removed.
Example 1:
Input
["LFUCache", "put", "put", "get", "put", "get", "get", "put", "get", "get", "get"]
[[2], [1, 1], [2, 2], [1], [3, 3], [2], [3], [4, 4], [1], [3], [4]]
Output
[null, null, null, 1, null, -1, 3, null, -1, 3, 4]
Explanation
LFUCache lfu = new LFUCache(2);
lfu.put(1, 1);
lfu.put(2, 2);
lfu.get(1); // return 1
lfu.put(3, 3); // evicts key 2
lfu.get(2); // return -1 (not found)
lfu.get(3); // return 3
lfu.put(4, 4); // evicts key 1.
lfu.get(1); // return -1 (not found)
lfu.get(3); // return 3
lfu.get(4); // return 4
Constraints:
0 <= capacity, key, value <= 104
- At most
10^5
calls will be made to get
and put
.
Follow up: Could you do both operations in O(1)
time complexity?
题目大意
请你为 最不经常使用(LFU)缓存算法设计并实现数据结构。
实现 LFUCache 类:
- LFUCache(int capacity) - 用数据结构的容量 capacity 初始化对象
- int get(int key) - 如果键存在于缓存中,则获取键的值,否则返回 -1。
- void put(int key, int value) - 如果键已存在,则变更其值;如果键不存在,请插入键值对。当缓存达到其容量时,则应该在插入新项之前,使最不经常使用的项无效。在此问题中,当存在平局(即两个或更多个键具有相同使用频率)时,应该去除 最久未使用 的键。
注意「项的使用次数」就是自插入该项以来对其调用 get 和 put 函数的次数之和。使用次数会在对应项被移除后置为 0 。
进阶:你是否可以在 O(1) 时间复杂度内执行两项操作?
解题思路
代码
package leetcode
import "container/list"
type LFUCache struct {
nodes map[int]*list.Element
lists map[int]*list.List
capacity int
min int
}
type node struct {
key int
value int
frequency int
}
func Constructor(capacity int) LFUCache {
return LFUCache{nodes: make(map[int]*list.Element),
lists: make(map[int]*list.List),
capacity: capacity,
min: 0,
}
}
func (this *LFUCache) Get(key int) int {
value, ok := this.nodes[key]
if !ok {
return -1
}
currentNode := value.Value.(*node)
this.lists[currentNode.frequency].Remove(value)
currentNode.frequency++
if _, ok := this.lists[currentNode.frequency]; !ok {
this.lists[currentNode.frequency] = list.New()
}
newList := this.lists[currentNode.frequency]
newNode := newList.PushBack(currentNode)
this.nodes[key] = newNode
if currentNode.frequency-1 == this.min && this.lists[currentNode.frequency-1].Len() == 0 {
this.min++
}
return currentNode.value
}
func (this *LFUCache) Put(key int, value int) {
if this.capacity == 0 {
return
}
if currentValue, ok := this.nodes[key]; ok {
currentNode := currentValue.Value.(*node)
currentNode.value = value
this.Get(key)
return
}
if this.capacity == len(this.nodes) {
currentList := this.lists[this.min]
frontNode := currentList.Front()
delete(this.nodes, frontNode.Value.(*node).key)
currentList.Remove(frontNode)
}
this.min = 1
currentNode := &node{
key: key,
value: value,
frequency: 1,
}
if _, ok := this.lists[1]; !ok {
this.lists[1] = list.New()
}
newList := this.lists[1]
newNode := newList.PushBack(currentNode)
this.nodes[key] = newNode
}