LRU 最近最少使用
设计可以变更的缓存结构（LRU）
【题目】
设计一种缓存结构，该结构在构造时确定大小，假设大小为K，并有两个功能：
set(key,value)：将记录(key,value)插入该结构。
get(key)：返回key对应的value值。
【要求】
1．set和get方法的时间复杂度为O(1)。
2．某个key的set或get操作一旦发生，认为这个key的记录成了最经常使用的。
3．当缓存的大小超过K时，移除最不经常使用的记录，即set或get最久远的。
【举例】
假设缓存结构的实例是cache，大小为3，并依次发生如下行为：
1．cache.set("A",1)。最经常使用的记录为("A",1)。
2．cache.set("B",2)。最经常使用的记录为("B",2)，("A",1)变为最不经常的。
3．cache.set("C",3)。最经常使用的记录为("C",2)，("A",1)还是最不经常的。
4．cache.get("A")。最经常使用的记录为("A",1)，("B",2)变为最不经常的。
5．cache.set("D",4)。大小超过了3，所以移除此时最不经常使用的记录("B",2)，
加入记录 ("D",4)，并且为最经常使用的记录，然后("C",2)变为最不经常使用的
记录
思路：
准备一个map和双向链表，map（A，node(A,3)）,key是原来的key，value是链表的node，那么就可以直接在双向链表中找到该node，O（1）。
public static class Node<V> {
public V value;
public Node<V> last;
public Node<V> next;
public Node(V value) {
this.value = value;
}
}
public static class NodeDoubleLinkedList<V> {
private Node<V> head;
private Node<V> tail;
public NodeDoubleLinkedList() {
this.head = null;
this.tail = null;
}
public void addNode(Node<V> newNode) {
if (newNode == null) {
return;
}
if (this.head == null) {
this.head = newNode;
this.tail = newNode;
} else {
this.tail.next = newNode;
newNode.last = this.tail;
this.tail = newNode;
}
}
public void moveNodeToTail(Node<V> node) {
if (this.tail == node) {
return;
}
if (this.head == node) {
this.head = node.next;
this.head.last = null;
} else {
node.last.next = node.next;
node.next.last = node.last;
}
node.last = this.tail;
node.next = null;
this.tail.next = node;
this.tail = node;
}
public Node<V> removeHead() {
if (this.head == null) {
return null;
}
Node<V> res = this.head;
if (this.head == this.tail) {
this.head = null;
this.tail = null;
} else {
this.head = res.next;
res.next = null;
this.head.last = null;
}
return res;
}
}
public static class MyCache<K, V> {
private HashMap<K, Node<V>> keyNodeMap;
private HashMap<Node<V>, K> nodeKeyMap;
private NodeDoubleLinkedList<V> nodeList;
private int capacity;
public MyCache(int capacity) {
if (capacity < 1) {
throw new RuntimeException("should be more than 0.");
}
this.keyNodeMap = new HashMap<K, Node<V>>();
this.nodeKeyMap = new HashMap<Node<V>, K>();
this.nodeList = new NodeDoubleLinkedList<V>();
this.capacity = capacity;
}
public V get(K key) {
if (this.keyNodeMap.containsKey(key)) {
Node<V> res = this.keyNodeMap.get(key);
this.nodeList.moveNodeToTail(res);
return res.value;
}
return null;
}
public void set(K key, V value) {
if (this.keyNodeMap.containsKey(key)) {
Node<V> node = this.keyNodeMap.get(key);
node.value = value;
this.nodeList.moveNodeToTail(node);
} else {
Node<V> newNode = new Node<V>(value);
this.keyNodeMap.put(key, newNode);
this.nodeKeyMap.put(newNode, key);
this.nodeList.addNode(newNode);
if (this.keyNodeMap.size() == this.capacity + 1) {
this.removeMostUnusedCache();
}
}
}
private void removeMostUnusedCache() {
Node<V> removeNode = this.nodeList.removeHead();
K removeKey = this.nodeKeyMap.get(removeNode);
this.nodeKeyMap.remove(removeNode);
this.keyNodeMap.remove(removeKey);
}
}
LFU 最近最不常使用
思路：维护一个双向链表，链表里是使用频率：1、2、3等，每个频率里挂一个双向链表，挂node。
image.png
增加一个节点：放在对应次数的头部
删除一个节点：删除对应次数的尾部（该次数下最早操作过的对象）
public class Code_03_LFU {
public static class Node {
public Integer key;
public Integer value;
public Integer times;
public Node up;
public Node down;
public Node(int key, int value, int times) {
this.key = key;
this.value = value;
this.times = times;
}
}
public static class LFUCache {
public static class NodeList {
public Node head;
public Node tail;
public NodeList last;
public NodeList next;
public NodeList(Node node) {
head = node;
tail = node;
}
public void addNodeFromHead(Node newHead) {
newHead.down = head;
head.up = newHead;
head = newHead;
}
public boolean isEmpty() {
return head == null;
}
public void deleteNode(Node node) {
if (head == tail) {
head = null;
tail = null;
} else {
if (node == head) {
head = node.down;
head.up = null;
} else if (node == tail) {
tail = node.up;
tail.down = null;
} else {
node.up.down = node.down;
node.down.up = node.up;
}
}
node.up = null;
node.down = null;
}
}
private int capacity;
private int size;
private HashMap<Integer, Node> records;
private HashMap<Node, NodeList> heads;
private NodeList headList;
public LFUCache(int capacity) {
this.capacity = capacity;
this.size = 0;
this.records = new HashMap<>();
this.heads = new HashMap<>();
headList = null;
}
public void set(int key, int value) {
if (records.containsKey(key)) {
Node node = records.get(key);
node.value = value;
node.times++;
NodeList curNodeList = heads.get(node);
move(node, curNodeList);
} else {
if (size == capacity) {
Node node = headList.tail;
headList.deleteNode(node);
modifyHeadList(headList);
records.remove(node.key);
heads.remove(node);
size--;
}
Node node = new Node(key, value, 1);
if (headList == null) {
headList = new NodeList(node);
} else {
if (headList.head.times.equals(node.times)) {
headList.addNodeFromHead(node);
} else {
NodeList newList = new NodeList(node);
newList.next = headList;
headList.last = newList;
headList = newList;
}
}
records.put(key, node);
heads.put(node, headList);
size++;
}
}
private void move(Node node, NodeList oldNodeList) {
oldNodeList.deleteNode(node);
NodeList preList = modifyHeadList(oldNodeList) ? oldNodeList.last
: oldNodeList;
NodeList nextList = oldNodeList.next;
if (nextList == null) {
NodeList newList = new NodeList(node);
if (preList != null) {
preList.next = newList;
}
newList.last = preList;
if (headList == null) {
headList = newList;
}
heads.put(node, newList);
} else {
if (nextList.head.times.equals(node.times)) {
nextList.addNodeFromHead(node);
heads.put(node, nextList);
} else {
NodeList newList = new NodeList(node);
if (preList != null) {
preList.next = newList;
}
newList.last = preList;
newList.next = nextList;
nextList.last = newList;
if (headList == nextList) {
headList = newList;
}
heads.put(node, newList);
}
}
}
// return whether delete this head
private boolean modifyHeadList(NodeList nodeList) {
if (nodeList.isEmpty()) {
if (headList == nodeList) {
headList = nodeList.next;
if (headList != null) {
headList.last = null;
}
} else {
nodeList.last.next = nodeList.next;
if (nodeList.next != null) {
nodeList.next.last = nodeList.last;
}
}
return true;
}
return false;
}
public int get(int key) {
if (!records.containsKey(key)) {
return -1;
}
Node node = records.get(key);
node.times++;
NodeList curNodeList = heads.get(node);
move(node, curNodeList);
return node.value;
}
}
}
