LeetCode – LRU Cache (Java)

Problem

Design and implement a data structure for Least Recently Used (LRU) cache. It should support the following operations: get and set.

get(key) - Get the value (will always be positive) of the key if the key exists in the cache, otherwise return -1.
set(key, value) - Set or insert the value if the key is not already present. When the cache reached its capacity, it should invalidate the least recently used item before inserting a new item.

Java Solution

The key to solve this problem is using a double linked list which enables us to quickly move nodes.

import java.util.HashMap;
 
public class LRUCache {
	private HashMap<Integer, DoubleLinkedListNode> map 
		= new HashMap<Integer, DoubleLinkedListNode>();
	private DoubleLinkedListNode head;
	private DoubleLinkedListNode end;
	private int capacity;
	private int len;
 
	public LRUCache(int capacity) {
		this.capacity = capacity;
		len = 0;
	}
 
	public int get(int key) {
		if (map.containsKey(key)) {
			DoubleLinkedListNode latest = map.get(key);
			removeNode(latest);
			setHead(latest);
			return latest.val;
		} else {
			return -1;
		}
	}
 
	public void removeNode(DoubleLinkedListNode node) {
		DoubleLinkedListNode cur = node;
		DoubleLinkedListNode pre = cur.pre;
		DoubleLinkedListNode post = cur.next;
 
		if (pre != null) {
			pre.next = post;
		} else {
			head = post;
		}
 
		if (post != null) {
			post.pre = pre;
		} else {
			end = pre;
		}
	}
 
	public void setHead(DoubleLinkedListNode node) {
		node.next = head;
		node.pre = null;
		if (head != null) {
			head.pre = node;
		}
 
		head = node;
		if (end == null) {
			end = node;
		}
	}
 
	public void set(int key, int value) {
		if (map.containsKey(key)) {
			DoubleLinkedListNode oldNode = map.get(key);
			oldNode.val = value;
			removeNode(oldNode);
			setHead(oldNode);
		} else {
			DoubleLinkedListNode newNode = 
				new DoubleLinkedListNode(key, value);
			if (len < capacity) {
				setHead(newNode);
				map.put(key, newNode);
				len++;
			} else {
				map.remove(end.key);
				end = end.pre;
				if (end != null) {
					end.next = null;
				}
 
				setHead(newNode);
				map.put(key, newNode);
			}
		}
	}
}
 
class DoubleLinkedListNode {
	public int val;
	public int key;
	public DoubleLinkedListNode pre;
	public DoubleLinkedListNode next;
 
	public DoubleLinkedListNode(int key, int value) {
		val = value;
		this.key = key;
	}
}

  

ps:

存在并发问题。

原文地址:https://www.cnblogs.com/huangfox/p/4316123.html