LinkedHashMap源码分析

LinkedHashMap源码分析

为什么要有LinkedHashMap?

在分析HashMap的时候提到了HashMap是无序的,即添加节点的顺序和遍历的顺序不一致

@Test
	public void test1() {
		HashMap<String,String> hashMap=new HashMap<String, String>();
		hashMap.put("tom", "american");
		hashMap.put("jack", "chainese");
		hashMap.put("mary", "japanese");
		Set<Entry<String, String>> entrySet = hashMap.entrySet();
		Iterator<Entry<String, String>> iterator = entrySet.iterator();
		while(iterator.hasNext()) {
			Entry<String, String> entry = iterator.next();
			String key = entry.getKey();
			String value = entry.getValue();
			System.out.println(key+":"+value);
		}
	}
输出:
tom:american
mary:japanese
jack:chainese

LinkedHashMap保证节点的顺序,这也是LinkedHashMap和HashMap的主要区别

@Test
	public void test2() {
		LinkedHashMap<String,String> linkedHashMap=new LinkedHashMap();
		linkedHashMap.put("tom", "american");
		linkedHashMap.put("jack", "chainese");
		linkedHashMap.put("mary", "japanese");
		Set<Entry<String, String>> entrySet = linkedHashMap.entrySet();
		Iterator<Entry<String, String>> iterator = entrySet.iterator();
		while(iterator.hasNext()) {
			Entry<String, String> entry = iterator.next();
			String key = entry.getKey();
			String value = entry.getValue();
			System.out.println(key+":"+value);
		}
	}
输出:
tom:american
jack:chainese
mary:japanese

存储示意图

类结构

public class LinkedHashMap<K,V>
    extends HashMap<K,V>
    implements Map<K,V>

LinkedHashMap是HashMap的子类,它对HashMap做了一些增强

节点

static class Entry<K,V> extends HashMap.Node<K,V> {//LinkedHashMap的Entry节点是HashMap的Node节点的子类
        Entry<K,V> before, after;//新增了before、after分别指向前驱和后继
        Entry(int hash, K key, V value, Node<K,V> next) {
          //直接使用HashMap的Node节点构造方法
            super(hash, key, value, next);
        }
    }

属性

//头指针指向第一个添加节点
transient LinkedHashMap.Entry<K,V> head;

//尾指针指向最后一个添加节点
transient LinkedHashMap.Entry<K,V> tail;

//排序规则,true的话按照访问顺序排序,最近访问的放到最后,false也是默认按照插入顺序排序
final boolean accessOrder;

构造方法

//指定初始化容量和加载因子
public LinkedHashMap(int initialCapacity, float loadFactor) {
        super(initialCapacity, loadFactor);
        accessOrder = false;
    }

//指定初始化容量
public LinkedHashMap(int initialCapacity) {
        super(initialCapacity);
        accessOrder = false;
    }

//无参构造方法
public LinkedHashMap() {
        super();
        accessOrder = false;
    }

//使用Map初始化
public LinkedHashMap(Map<? extends K, ? extends V> m) {
        super();
        accessOrder = false;
        putMapEntries(m, false);
    }

//指定初始化容量、加载因子、排序规则
public LinkedHashMap(int initialCapacity,
                         float loadFactor,
                         boolean accessOrder) {
        super(initialCapacity, loadFactor);
        this.accessOrder = accessOrder;
    }

方法

LinkedHashMap中并没有put方法,所以使用的是父类HashMap的put方法

public V put(K key, V value) {
        return putVal(hash(key), key, value, false, true);
    }

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)
          //当根据hash计算的下标位置没放节点,调用LinkedHashMap的newNode方法
            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;
              //如果是指定访问顺序排序,那么替换后,把节点移动到最后
                afterNodeAccess(e);
                return oldValue;
            }
        }
        ++modCount;
        if (++size > threshold)
            resize();
        afterNodeInsertion(evict);
        return null;
    }

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) {
  //保存LinkedHashMap的为指针
        LinkedHashMap.Entry<K,V> last = tail;
        tail = p;
        if (last == null)//尾指针为null说明LinkedHashMap中没元素
            head = p;
        else {//有元素
          //新节点的前驱指向添加之前的尾指针
            p.before = last;
          //添加之前的尾指针节点的后继指向新节点
            last.after = p;
        }
    }

void afterNodeAccess(Node<K,V> e) { // move node to last
        LinkedHashMap.Entry<K,V> last;
        if (accessOrder && (last = tail) != e) {//如果指定了按访问顺序排序且替换的节点不是最末尾的节点
            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;
        }
  
  void afterNodeInsertion(boolean evict) { // possibly remove eldest
        LinkedHashMap.Entry<K,V> first;
        if (evict && (first = head) != null && removeEldestEntry(first)) {//removeEldestEntry方法返回false所以不会进入if
            K key = first.key;
            removeNode(hash(key), key, null, false, true);
        }
    }

get(Object)根据key获取值

public V get(Object key) {
        Node<K,V> e;
        if ((e = getNode(hash(key), key)) == null)//调用HashMap的方法拿到值
            return null;
        if (accessOrder)//如果是按照访问顺序排序的话
          //访问过后要修改顺序
            afterNodeAccess(e);
        return e.value;
    }

//把访问的节点移动到链表的最末端
void afterNodeAccess(Node<K,V> e) { 
        LinkedHashMap.Entry<K,V> last;
        if (accessOrder && (last = tail) != e) {//按访问顺序排序并且访问节点不是最后一个节点
            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;
        }
    }

remove(Object)方法是调用父类HashMap的方法

public V remove(Object key) {
        Node<K,V> e;
        return (e = removeNode(hash(key), key, null, false, true)) == null ?
            null : e.value;
    }

 final Node<K,V> removeNode(int hash, Object key, Object value,
                               boolean matchValue, boolean movable) {
        Node<K,V>[] tab; Node<K,V> p; int n, index;
        if ((tab = table) != null && (n = tab.length) > 0 &&
            (p = tab[index = (n - 1) & hash]) != null) {
            Node<K,V> node = null, e; K k; V v;
            if (p.hash == hash &&
                ((k = p.key) == key || (key != null && key.equals(k))))
                node = p;
            else if ((e = p.next) != null) {
                if (p instanceof TreeNode)
                    node = ((TreeNode<K,V>)p).getTreeNode(hash, key);
                else {
                    do {
                        if (e.hash == hash &&
                            ((k = e.key) == key ||
                             (key != null && key.equals(k)))) {
                            node = e;
                            break;
                        }
                        p = e;
                    } while ((e = e.next) != null);
                }
            }
            if (node != null && (!matchValue || (v = node.value) == value ||
                                 (value != null && value.equals(v)))) {
                if (node instanceof TreeNode)
                    ((TreeNode<K,V>)node).removeTreeNode(this, tab, movable);
                else if (node == p)
                    tab[index] = node.next;
                else
                    p.next = node.next;
                ++modCount;
                --size;
              //调用LinkedHashMap·的方法来实现双链的删除
                afterNodeRemoval(node);
                return node;
            }
        }
        return null;
    }

void afterNodeRemoval(Node<K,V> e) { // unlink
        LinkedHashMap.Entry<K,V> p =
            (LinkedHashMap.Entry<K,V>)e, b = p.before, a = p.after;
  //把要移除节点的前驱后继置为null
        p.before = p.after = null;
        if (b == null)//b为null即移除的就是第一个元素
          //头指针指向移除元素的后继
            head = a;
        else
            b.after = a;
        if (a == null)//a为null即移除的元素是最后一个元素
          //尾指针指向移除元素的前驱
            tail = b;
        else
            a.before = b;
    }
posted @ 2020-04-23 20:59  moyuduo  阅读(270)  评论(0编辑  收藏  举报