460. LFU Cache #

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 #

这一题是 LFU 经典面试题，详细解释见第三章模板。

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
}

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