礼品公司怎么做网站aso优化技术

Least Frequently Used（最不频繁使用）

Leetcode有原题，之前手写过LRU，数据结构还是习惯于用java实现，实现是copy的评论题解。
题解注释写的很清楚

大致就是说LFUCache类维护一个存放node的map，同时维护两个双向链表，注意这个双向链表里面又包含了两个双向链表，访问的频率是first最大，last最小。其余的就是正常的双向链表的操作了(插入，删除)

import java.util.HashMap;
import java.util.Map;class LFUCache {Map<Integer, Node> cache; // 存储缓存的内容，Node中除了value值外，还有key、freq、所在doublyLinkedList、所在doublyLinkedList中的postNode、所在doublyLinkedList中的preNode，具体定义在下方。DoublyLinkedList firstLinkedList; // firstLinkedList.post 是频次最大的双向链表DoublyLinkedList lastLinkedList; // lastLinkedList.pre 是频次最小的双向链表，满了之后删除 lastLinkedList.pre.tail.pre// 这个Node即为频次最小且访问最早的Nodeint size;int capacity;public LFUCache(int capacity) {cache = new HashMap<>(capacity);firstLinkedList = new DoublyLinkedList();lastLinkedList = new DoublyLinkedList();firstLinkedList.post = lastLinkedList;  // 是按照倒序排列的，最大的是firstlastLinkedList.pre = firstLinkedList;this.capacity = capacity;}public int get(int key) {Node node = cache.get(key);if (node == null) {return -1;}// 该key访问频次+1freqInc(node);return node.value;}public void put(int key, int value) {if (capacity == 0) {return;}Node node = cache.get(key);// 若key存在，则更新value，访问频次+1if (node != null) {node.value = value;freqInc(node);} else {// 若key不存在if (size == capacity) {// 如果缓存满了，删除lastLinkedList.pre这个链表（即表示最小频次的链表）中的tail.pre这个Node（即最小频次链表中最先访问的Node），如果该链表中的元素删空了，则删掉该链表。cache.remove(lastLinkedList.pre.tail.pre.key);lastLinkedList.removeNode(lastLinkedList.pre.tail.pre);size--;if (lastLinkedList.pre.head.post == lastLinkedList.pre.tail) {removeDoublyLinkedList(lastLinkedList.pre);}}// cache中put新Key-Node对儿，并将新node加入表示freq为1的DoublyLinkedList中，若不存在freq为1的DoublyLinkedList则新建。Node newNode = new Node(key, value);cache.put(key, newNode);if (lastLinkedList.pre.freq != 1) {DoublyLinkedList newDoublyLinedList = new DoublyLinkedList(1);addDoublyLinkedList(newDoublyLinedList, lastLinkedList.pre);newDoublyLinedList.addNode(newNode);} else {lastLinkedList.pre.addNode(newNode);}size++;}}/*** node的访问频次 + 1*/void freqInc(Node node) {// 将node从原freq对应的双向链表里移除, 如果链表空了则删除链表。DoublyLinkedList linkedList = node.doublyLinkedList;DoublyLinkedList preLinkedList = linkedList.pre;linkedList.removeNode(node);if (linkedList.head.post == linkedList.tail) {removeDoublyLinkedList(linkedList);}// 将node加入新freq对应的双向链表，若该链表不存在，则先创建该链表。node.freq++;if (preLinkedList.freq != node.freq) {DoublyLinkedList newDoublyLinedList = new DoublyLinkedList(node.freq);addDoublyLinkedList(newDoublyLinedList, preLinkedList);newDoublyLinedList.addNode(node);} else {preLinkedList.addNode(node);}}/*** 增加代表某1频次的双向链表*/void addDoublyLinkedList(DoublyLinkedList newDoublyLinedList, DoublyLinkedList preLinkedList) {newDoublyLinedList.post = preLinkedList.post;newDoublyLinedList.post.pre = newDoublyLinedList;newDoublyLinedList.pre = preLinkedList;preLinkedList.post = newDoublyLinedList;}/*** 删除代表某1频次的双向链表*/void removeDoublyLinkedList(DoublyLinkedList doublyLinkedList) {doublyLinkedList.pre.post = doublyLinkedList.post;doublyLinkedList.post.pre = doublyLinkedList.pre;}}class Node {int key;int value;int freq = 1;Node pre; // Node所在频次的双向链表的前继NodeNode post; // Node所在频次的双向链表的后继NodeDoublyLinkedList doublyLinkedList; // Node所在频次的双向链表public Node() {}public Node(int key, int value) {this.key = key;this.value = value;}}class DoublyLinkedList {int freq; // 该双向链表表示的频次DoublyLinkedList pre; // 该双向链表的前继链表（pre.freq < this.freq）DoublyLinkedList post; // 该双向链表的后继链表 (post.freq > this.freq)Node head; // 该双向链表的头节点，新节点从头部加入，表示最近访问Node tail; // 该双向链表的尾节点，删除节点从尾部删除，表示最久访问public DoublyLinkedList() {head = new Node();tail = new Node();head.post = tail;tail.pre = head;}public DoublyLinkedList(int freq) {head = new Node();tail = new Node();head.post = tail;tail.pre = head;this.freq = freq;}void removeNode(Node node) {node.pre.post = node.post;node.post.pre = node.pre;}void addNode(Node node) {node.post = head.post;head.post.pre = node;head.post = node;node.pre = head;node.doublyLinkedList = this;}}