面试被要求手写LRU 缓存机制算法，使用LinkedHashMap实现

叁歲伎倆 2022-08-28 05:53 132阅读 0赞

方式2:面试官说，不允许使用LinkedHashMap，请你手写一个。

> 你是否可以在 `O(1)` 时间复杂度内完成这两种操作？

这是力扣题,官网地址:[146. LRU 缓存机制][146. LRU]

public class LRUCache_2 { 
        /** * map负责查找，构建一个虚拟的双向链表，它里面安装的就是一个Node节点，作为数据载体 * 1，构建一个Node节点，作为数据载体 * @param <K> * @param <V> */
        class Node<K,V> { 
            K key;
            V value;
            Node<K,V> prev; // 前缀
            Node<K,V> next; // 后缀
            public Node() { 
                // 将前后缀初始化为null
                this.prev = this.next = null;
            }
            public Node(K key, V value) { 
                this.key = key;
                this.value = value;
                this.prev = this.next = null;
            }
        }
    
        // 构建一个双向队列，里面安装就是我们的Node
        class DoubleLinkedList<K,V> { 
            Node<K,V> head; //头节点
            Node<K,V> tail; //尾节点
            public DoubleLinkedList() { 
                head = new Node<>();
                tail = new Node<>();
                // 头尾连接构成双向链表
                head.next = tail;
                tail.prev = head;
            }
            // 添加到头
            public void addHead(Node<K,V> node) { 
                node.next = head.next;
                node.prev = head;
                head.next.prev = node;
                head.next = node;
            }
            // 删除节点
            public void removeNode(Node<K,V> node) { 
                node.next.prev = node.prev;
                node.prev.next = node.next;
                node.prev = null;
                node.next = null;
            }
            // 获取
            public Node getLast() { 
                return tail.prev;
            }
        }
        private int cacheSize;
        Map<Integer, Node<Integer,Integer>> map;
        DoubleLinkedList<Integer,Integer> doubleLinkedList;
    
        public LRUCache_2(int cacheSize) { 
            this.cacheSize = cacheSize;
            map = new HashMap<>();
            doubleLinkedList = new DoubleLinkedList<>();
        }
        public int get(int key) { 
            if (!map.containsKey(key)) { 
                // 如果不存在就返回-1
                return -1;
            }
            Node<Integer, Integer> node = map.get(key);
            doubleLinkedList.removeNode(node);
            doubleLinkedList.addHead(node);
            return node.value;
        }
        public void put(int key, int value) { 
            if (map.containsKey(key)) { 
                Node<Integer, Integer> node = map.get(key);
                node.value = value;
                map.put(key,node);
                doubleLinkedList.removeNode(node);
                doubleLinkedList.addHead(node);
            } else { 
                if (map.size() == cacheSize) { 
                    Node<Integer,Integer> last = doubleLinkedList.getLast();
                    map.remove(last.key);
                    doubleLinkedList.removeNode(last);
                }
                // 新增
                Node<Integer, Integer> newNode = new Node<>(key, value);
                map.put(key,newNode);
                doubleLinkedList.addHead(newNode);
            }
        }
        public static void main(String[] args) { 
            LRUCache_2 lruCache = new LRUCache_2(3);
            lruCache.put(1,1);
            lruCache.put(2,2);
            lruCache.put(3,3);
            System.out.println(lruCache.map.keySet());
            lruCache.put(4,4);
            System.out.println(lruCache.map.keySet());
        }
    }

更多文章收录于[https://github.com/niutongg/JavaLeague][https_github.com_niutongg_JavaLeague]

[146. LRU]: https://leetcode-cn.com/problems/lru-cache/
[https_github.com_niutongg_JavaLeague]: https://github.com/niutongg/JavaLeague