使用HashMap+双向链表实现LRU

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

get(key) - Get the value (will always be positive) of the key if the key exists in the cache, otherwise return -1.
put(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.

The cache is initialized with a positive capacity.

Follow up:
Could you do both operations in O(1) time complexity?

Example:

LRUCache cache = new LRUCache( 2 /* capacity */ );

cache.put(1, 1);
cache.put(2, 2);
cache.get(1); // returns 1
cache.put(3, 3); // evicts key 2
cache.get(2); // returns -1 (not found)
cache.put(4, 4); // evicts key 1
cache.get(1); // returns -1 (not found)
cache.get(3); // returns 3
cache.get(4); // returns 4

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思路：

可以直接使用LinkHashMap实现LRU。

LinkedHashMap实现原理基于HashMap+双向链表，这里自己实现一遍。

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题解：

import java.util.HashMap;

class LRUCache {

    class Node {
        int key;
        int value;
        Node prev;
        Node next;
    }

    /**
     * 向双向链表加入新节点
     * 加入到表头
     */
    private void addNode(Node node) {
        node.prev = head;
        node.next = head.next;
        head.next.prev = node;
        head.next = node;
    }

    /**
     * 删除节点
     */
    private void removeNode(Node node) {
        node.prev.next = node.next;
        node.next.prev = node.prev;
    }

    /**
     * 删除表尾节点
     */
    private Node popTail() {
        Node node = tail.prev;
        removeNode(node);
        return node;
    }

    /**
     * 移动到表头
     * 1.删除原节点
     * 2.在表头插入原节点
     */
    private void moveToHead(Node node) {
        removeNode(node);
        addNode(node);
    }

    private int capacity;
    private int size;
    private HashMap cache;
    private Node head;
    private Node tail;

    public LRUCache(int capacity) {
        this.capacity = capacity;
        this.size = 0;
        cache = new HashMap<>();
        head = new Node();
        tail = new Node();
        head.next = tail;
        tail.prev = head;
    }

    /**
     * get()操作
     * 如果key对应的value不存在，返回-1
     * 如果存在，则获取值，并将节点移动到表头
     */
    public int get(int key) {
        Node node = cache.get(key);
        if (node == null) return -1;
        moveToHead(node);
        return node.value;
    }

    /**
     * put()操作
     * 如果key不存在，则新建节点，将节点加入cache map，放置到表头，size+1
     * 如果size>capacity，要去除表尾节点，并从cache中删除，size-1
     * 如果key存在，将该节点的值更新，并移动到表头
     */
    public void put(int key, int value) {
        Node node = cache.get(key);
        if (node == null) {
            node = new Node();
            node.key = key;
            node.value = value;
            cache.put(key, node);
            addNode(node);
            size++;
            if (size > capacity) {
                Node del = popTail();
                cache.remove(del.key);
                size--;
            }
        } else {
            node.value = value;
            moveToHead(node);
        }
    }

    public static void main(String[] args) {
        LRUCache lruCache = new LRUCache(2);
        lruCache.put(1, 1);
        lruCache.put(2, 2);
        lruCache.get(1);
        lruCache.put(3, 3);
        lruCache.get(2);
        lruCache.put(4, 4);
        lruCache.get(1);
        lruCache.get(3);
        lruCache.get(4);
    }
}

/**
 * Your LRUCache object will be instantiated and called as such:
 * LRUCache obj = new LRUCache(capacity);
 * int param_1 = obj.get(key);
 * obj.put(key,value);
 */

 

转载于:https://www.cnblogs.com/yfzhou/p/11246472.html