LinkedHashMap

文章目录

      • LinkedHashMap详解
        • 基本数据结构
        • 遍历顺序按照插入顺序
        • 遍历顺序按照访问顺序
        • LinkedHashMap实现LRU

LinkedHashMap详解

相信大家都知道,HashMap的访问时无序的,如果你想按照插入顺序,访问元素,就必须使用

LinkedHashMap, 下面详细讲述其原理。

基本数据结构

LinkedHashMap 的大致数据结构如下图所示

(链表和哈希表中相同的键值对都是指向同一个对象,区分开来是为了呈现清晰的结构)

其中双向链表,对元素进行排序,保证按照插入的顺序,访问元素

LinkedHashMap_第1张图片

遍历顺序按照插入顺序

LinkedHashMap的链表节点继承了HashMap的节点,而且每个节点都包含了前指针和后指针,所以这里可以看出它是一个双向链表

static class Entry<K,V> extends HashMap.Node<K,V> {
     
    Entry<K,V> before, after;
    Entry(int hash, K key, V value, Node<K,V> next) {
     
    super(hash, key, value, next);
    }
}

头指针和尾指针

/**
* The head (eldest) of the doubly linked list.
*/
transient LinkedHashMap.Entry<K,V> head;
/**
* The tail (youngest) of the doubly linked list.
*/
transient LinkedHashMap.Entry<K,V> tail;

在插入key,value键值对后,如果是新的键值对(不存在重复key),会将此键值对作为双链表的尾节点

调用put方法后,如果是新的键值对,一定会调用newNode方法

final V putVal(int hash, K key, V value, boolean onlyIfAbsent,
boolean evict) {
     
    Node<K,V>[] tab; Node<K,V> p; int n, i;
    if ((tab = table) == null || (n = tab.length) == 0)
    	n = (tab = resize()).length;
    if ((p = tab[i = (n - 1) & hash]) == null)
    	tab[i] = newNode(hash, key, value, null);
    else {
     
        Node<K,V> e; K k;
        if (p.hash == hash &&
        ((k = p.key) == key || (key != null && key.equals(k))))
        	e = p;
        else if (p instanceof TreeNode)
        	e = ((TreeNode<K,V>)p).putTreeVal(this, tab, hash, key, value);
        else {
     
            for (int binCount = 0; ; ++binCount) {
     
                if ((e = p.next) == null) {
     
                    p.next = newNode(hash, key, value, null);
                    if (binCount >= TREEIFY_THRESHOLD - 1) // -1 for 1st
                        treeifyBin(tab, hash);
                   		break;
                }
                if (e.hash == hash &&
                    ((k = e.key) == key || (key != null && key.equals(k))))
                   	 break;
                p = e;
            }
        }
        if (e != null) {
      // existing mapping for key
            V oldValue = e.value;
            if (!onlyIfAbsent || oldValue == null)
            	e.value = value;
            afterNodeAccess(e);
            return oldValue;
        }
    }
    ++modCount;
    if (++size > threshold)
    	resize();
    afterNodeInsertion(evict);
    return null;
}

newNode方法, 将新的键值对,作为双链表尾部节点

Node<K,V> newNode(int hash, K key, V value, Node<K,V> e) {
     
    LinkedHashMap.Entry<K,V> p =
        new LinkedHashMap.Entry<K,V>(hash, key, value, e);
    linkNodeLast(p);
    return p;
}

private void linkNodeLast(LinkedHashMap.Entry<K,V> p) {
     
    LinkedHashMap.Entry<K,V> last = tail;
    tail = p;
    if (last == null)
        head = p;
    else {
     
        p.before = last;
        last.after = p;
    }
}

遍历顺序按照访问顺序

final boolean accessOrder;

LinkedHashMap,默认是按照插入顺序,来遍历元素。但是同时也支持按照访问顺序(get方法),来遍历元素

遍历顺序按照访问顺序,只需要设置accessOrder =true

按照访问顺序遍历,举个具体的例子

插入,key为12345的键值对
执行一次get(2)的操作
遍历输出的结果为 1 3 4 5 2

如果设置accessOrder为true, LinkedMap在get元素后,会将元素移动至双链表的尾节点

执行get方法后,会执行afterNodeAcess方法

public V get(Object key) {
     
    Node<K,V> e;
    if ((e = getNode(hash(key), key)) == null)
    	return null;
    if (accessOrder)
    	afterNodeAccess(e);
    return e.value;
}

afterNodeAccess方法。将元素移动至双链表尾节点

void afterNodeAccess(Node<K,V> e) {
      // move node to last
    LinkedHashMap.Entry<K,V> last;
    if (accessOrder && (last = tail) != e) {
     
        LinkedHashMap.Entry<K,V> p =
        (LinkedHashMap.Entry<K,V>)e, b = p.before, a = p.after;
        p.after = null;
        if (b == null)
        	head = a;
        else
        	b.after = a;
        if (a != null)
        	a.before = b;
        else
        last = b;
        if (last == null)
        	head = p;
        else {
     
            p.before = last;
            last.after = p;
        }
        tail = p;
        ++modCount;
    }
}

需要注意的时,put时,发生值覆盖情况,也会支持afterNodeAccess方法

LinkedHashMap实现LRU

实现LRU要保证几点

  1. 插入时,新元素在双链表尾部,如果size超过数量,需要删除双链表头部元素(重写removeEldestEntry方法)
  2. 插入时,发生key值重复,即值覆盖情况,需要将相应元素移动至双链表尾部(accessOrder设置为true)
  3. 访问时,需要将相应的元素移动至队列尾部(accessOrder设置为true)
  4. 删除时,哈希表和双链表均需要删除相应元素(满足)

Leetcode真题https://leetcode-cn.com/problems/lru-cache-lcci/submissions

import java.util.*;

class ZYMap<K,V> extends LinkedHashMap<K,V> {
     
  private int cacheSize;
  
  public ZYMap(int cacheSize) {
     
      super(16,0.75f,true);
      this.cacheSize = cacheSize;
  }

  @Override
  protected boolean removeEldestEntry(Map.Entry<K, V> eldest) {
     
      return this.size() > cacheSize;
  }
}

class LRUCache {
     
    ZYMap map;
    public LRUCache(int capacity) {
     
        map = new ZYMap<Integer, Integer>(capacity);
    }
    
    public int get(int key) {
     
        if(map.get(key) == null) return -1;
        else return (int)map.get(key);
    }
    
    public void put(int key, int value) {
     
        map.put(key,value);
    }
}

/**
 * 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);
 */

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