手撕LRU缓存Java版(带输入输出)

由于面试手撕lru没撕出来,导致心态炸裂,今天特地练习了lru输入输出

手撕版,在每个函数里手动加上输出

java 复制代码
public class LC146 {
    static class LRUCache{
        class Node{
            int key, value;
            Node prev, next;
            Node(int key, int value){
                this.key = key;
                this.value = value;
            }
        }
        private Node dummy = new Node(0,0);
        private int capacity;

        private Map<Integer, Node> map = new HashMap<Integer, Node>();
        public LRUCache(int capacity){
            this.capacity = capacity;
            dummy.prev = dummy;
            dummy.next = dummy;
            System.out.print("null,");
        }

        public int get(int key) {
            Node node = getNode(key);
            if(node == null){
                System.out.print("-1,");
                return -1;
            }else{
                System.out.print(node.value+",");
                return node.value;
            }
        }

        private Node getNode(int key) {
            Node node = map.getOrDefault(key, null);
            if(node == null){
                return null;
            }
            deleteNode(node);
            pushFront(node);
            return node;
        }

        private void deleteNode(Node node) {
            node.next.prev = node.prev;
            node.prev.next = node.next;
        }

        private void pushFront(Node node) {
            node.next = dummy.next;
            node.prev = dummy;
            node.prev.next = node;
            node.next.prev = node;
        }

        public void put(int key, int value) {
            Node node = getNode(key);
            if(node != null){
                node.value = value;
                return ;
            }
            node = new Node(key, value);
            pushFront(node);
            map.put(key, node);
            if(map.size() > capacity){
                map.remove(dummy.prev.key);
                deleteNode(dummy.prev);
            }
            System.out.print("null,");
        }
    }

    public static void main(String[] args) {
        LRUCache lRUCache = new LRUCache(2);
        lRUCache.put(1, 1); // 缓存是 {1=1}
        lRUCache.put(2, 2); // 缓存是 {1=1, 2=2}
        lRUCache.get(1);    // 返回 1
        lRUCache.put(3, 3); // 该操作会使得关键字 2 作废,缓存是 {1=1, 3=3}
        lRUCache.get(2);    // 返回 -1 (未找到)
        lRUCache.put(4, 4); // 该操作会使得关键字 1 作废,缓存是 {4=4, 3=3}
        lRUCache.get(1);    // 返回 -1 (未找到)
        lRUCache.get(3);    // 返回 3
        lRUCache.get(4);    // 返回 4
    }
}

leetcode输入输出格式版

java 复制代码
public class LC146 {
    static class LRUCache{
        class Node{
            int key, value;
            Node prev, next;
            Node(int key, int value){
                this.key = key;
                this.value = value;
            }
        }
        private Node dummy = new Node(0,0);
        private int capacity;

        private Map<Integer, Node> map = new HashMap<Integer, Node>();
        public LRUCache(int capacity){
            this.capacity = capacity;
            dummy.prev = dummy;
            dummy.next = dummy;
            System.out.print("null,");
        }

        public int get(int key) {
            Node node = getNode(key);
            if(node == null){
                System.out.print("-1,");
                return -1;
            }else{
                System.out.print(node.value+",");
                return node.value;
            }
        }

        private Node getNode(int key) {
            Node node = map.getOrDefault(key, null);
            if(node == null){
                return null;
            }
            deleteNode(node);
            pushFront(node);
            return node;
        }

        private void deleteNode(Node node) {
            node.next.prev = node.prev;
            node.prev.next = node.next;
        }

        private void pushFront(Node node) {
            node.next = dummy.next;
            node.prev = dummy;
            node.prev.next = node;
            node.next.prev = node;
        }

        public void put(int key, int value) {
            Node node = getNode(key);
            if(node != null){
                node.value = value;
                return ;
            }
            node = new Node(key, value);
            pushFront(node);
            map.put(key, node);
            if(map.size() > capacity){
                map.remove(dummy.prev.key);
                deleteNode(dummy.prev);
            }
            System.out.print("null,");
        }
    }

    public static void main(String[] args) {
        String[] s = new String[]{"LRUCache", "put", "put", "get", "put", "get", "put", "get", "get", "get"};
        int[][] inputs = new int[][]{{2}, {1, 1}, {2, 2}, {1}, {3, 3}, {2}, {4, 4}, {1}, {3}, {4}};
        int n = s.length;
        LRUCache lruCache = null;
        for(int i=0; i<n; i++){
            if(s[i].equals("LRUCache")){
                lruCache = new LRUCache(inputs[i][0]);
                System.out.print("null");
            }else if(s[i].equals("put")){
                lruCache.put(inputs[i][0], inputs[i][1]);
                System.out.print("null");
            }else if(s[i].equals("get")){
                int ans = lruCache.get(inputs[i][0]);
                System.out.print(ans);
            }
            if(i != n-1){
                System.out.print(",");
            }
        }
    }
}

测试结果

相关推荐
JH30736 小时前
SpringBoot 优雅处理金额格式化:拦截器+自定义注解方案
java·spring boot·spring
Coder_Boy_7 小时前
技术让开发更轻松的底层矛盾
java·大数据·数据库·人工智能·深度学习
invicinble8 小时前
对tomcat的提供的功能与底层拓扑结构与实现机制的理解
java·tomcat
较真的菜鸟8 小时前
使用ASM和agent监控属性变化
java
黎雁·泠崖8 小时前
【魔法森林冒险】5/14 Allen类(三):任务进度与状态管理
java·开发语言
qq_12498707539 小时前
基于SSM的动物保护系统的设计与实现(源码+论文+部署+安装)
java·数据库·spring boot·毕业设计·ssm·计算机毕业设计
Coder_Boy_9 小时前
基于SpringAI的在线考试系统-考试系统开发流程案例
java·数据库·人工智能·spring boot·后端
Mr_sun.9 小时前
Day06——权限认证-项目集成
java
瑶山9 小时前
Spring Cloud微服务搭建四、集成RocketMQ消息队列
java·spring cloud·微服务·rocketmq·dashboard
abluckyboy9 小时前
Java 实现求 n 的 n^n 次方的最后一位数字
java·python·算法