題目
設計一種緩存結構,該結構在構造時確定大小,假設大小爲K,並由兩個功能:
- set(key,value):將記錄(key,value)插入該結構
- get(key):返回key對應的value值
要求
1.set和get方法的時間複雜度O(1)。
2.某個key的set或get操作一旦發生,認爲這個key的記錄成了最經常使用的。
3.當緩存的大小超過K時,移除最不經常使用的記錄,即set或get最久遠的。
JAVA代碼實現
package otherquestions;
/**
* 雙向鏈表節點的結構
* @author HDian
*
* @param <V>
*/
public class Node<V> {
public V value;
public Node<V> last;
public Node<V> next;
public Node(V value) {
this.value = value;
}
}
package otherquestions;
/**
* 基於雙向鏈表節點實現雙向鏈表,實現按照“訪問經常度”排序節點
* @author HDian
*
* @param <V>
*/
public class NodeDoubleLinkedList1217<V> {
//定義雙向鏈表的頭節點和尾節點
private Node<V> head;
private Node<V> tail;
/*
* 構造方法重載
*/
public NodeDoubleLinkedList1217() {
this.head = null;
this.tail = null;
}
/*
* 雙向鏈表中添加新節點
*/
public void addNode(Node<V> newNode) {
if (newNode == null) {
return;
}
if (this.head == null) {
this.head = newNode;
this.tail = newNode;
} else {
this.head.next = newNode;
newNode.last = this.tail;
this.tail = newNode;
}
}
/*
* 最常用的節點放到尾節點處
*/
public void moveNodeToTail(Node<V> node) {
if (this.tail == node) {
return;
}
if (this.head == node) {
this.head = node.next;
node.next.last = node.last;
}
node.last = this.tail;
node.next = null;
this.tail.next = node;
this.tail = node;
}
/*
* 頭節點刪除
*/
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;
}
}
package otherquestions;
import java.util.HashMap;
public class MyCache1217<K, V> {
//key到node的映射
private HashMap<K, Node<V>> keyNodeMap;
//node到key的映射
private HashMap<Node<V>, K> nodeKeyMap;
private NodeDoubleLinkedList1217<V> nodeList;
private int capacity;
//構造方法重載
public MyCache1217(int capacity) {
if (capacity < 1) {
throw new RuntimeException("should be more than 0.");
}
this.keyNodeMap = new HashMap<>();
this.nodeKeyMap = new HashMap<>();
this.nodeList = new NodeDoubleLinkedList1217<>();
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(removeKey);
this.keyNodeMap.remove(removeKey);
}
}