HashMap源码实现解析 1 2 3 java version "1.8.0_251" Java(TM) SE Runtime Environment (build 1.8.0_251-b08) Java HotSpot(TM) 64-Bit Server VM (build 25.251-b08, mixed mode)
基于数组+链表实现,通过&
与运算,计算数组下标。在JDK8中,加入红黑树实现,使其时间复杂度保持在O(1)到O(logn)
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 transient Node<K,V>[] table;static class Node <K,V> implements Map .Entry<K,V> { final int hash; final K key; V value; Node<K,V> next; Node(int hash, K key, V value, Node<K,V> next) { this .hash = hash; this .key = key; this .value = value; this .next = next; } public final K getKey () { return key; } public final V getValue () { return value; } public final String toString () { return key + "=" + value; } public final int hashCode () { return Objects.hashCode(key) ^ Objects.hashCode(value); } public final V setValue (V newValue) { V oldValue = value; value = newValue; return oldValue; } public final boolean equals (Object o) { if (o == this ) return true ; if (o instanceof Map.Entry) { Map.Entry<?,?> e = (Map.Entry<?,?>)o; if (Objects.equals(key, e.getKey()) && Objects.equals(value, e.getValue())) return true ; } return false ; } }
HashMap 静态内部类Node,实现链表,通过Node[]这个数组属性存放所有的节点。
put(K,V) 应该直接看final V putVal(int hash, K key, V value, boolean onlyIfAbsent, boolean evict)
这个方法更为实际
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 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 ) treeifyBin(tab, hash); break ; } if (e.hash == hash && ((k = e.key) == key || (key != null && key.equals(k)))) break ; p = e; } } if (e != null ) { V oldValue = e.value; if (!onlyIfAbsent || oldValue == null ) e.value = value; afterNodeAccess(e); return oldValue; } } ++modCount; if (++size > threshold) resize(); afterNodeInsertion(evict); return null ; }
如果当前想要存放的这个节点的hash值暂时没有存在的节点,则直接在数组中添加。
1 2 if ((p = tab[i = (n - 1 ) & hash]) == null ) tab[i] = newNode(hash, key, value, null );
通过&
与运算,
如果当前节点的hash值存在了,则在这个节点下增加链表。
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 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 ) treeifyBin(tab, hash); break ; } if (e.hash == hash && ((k = e.key) == key || (key != null && key.equals(k)))) break ; p = e; } } if (e != null ) { V oldValue = e.value; if (!onlyIfAbsent || oldValue == null ) e.value = value; afterNodeAccess(e); return oldValue; } }
从JDK8 开始,当链表中的子节点超过八个时,将转为红黑树。关于红黑树的数据结构特点,我现在也不是特别的理解,先给自己挖个坑,改天填。
红黑树 HashMap扩容 HashMap中有一个属性:threshold
,这个主要是根据阀值和当前HashMap的大小计算而来,可通过查看
1 2 3 4 5 6 7 8 9 10 11 12 public HashMap (int initialCapacity, float loadFactor) { if (initialCapacity < 0 ) throw new IllegalArgumentException ("Illegal initial capacity: " + initialCapacity); if (initialCapacity > MAXIMUM_CAPACITY) initialCapacity = MAXIMUM_CAPACITY; if (loadFactor <= 0 || Float.isNaN(loadFactor)) throw new IllegalArgumentException ("Illegal load factor: " + loadFactor); this .loadFactor = loadFactor; this .threshold = tableSizeFor(initialCapacity); }
在初始化HashMap的最后,会根据当前的阀值和实际的大小进行计算threshold
的值,同时在每一次操作元素的时候,都会去比较当前HashMap的实际大小与threshold
的值,如果当前实际大小已经大于了这个限定的阀值,此时将会对HashMap进行扩容。
resize()
方法主要是两个步骤:
计算大小;
将原HashMap中的元素进行移动
挖坑,以后填
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 int oldCap = (oldTab == null ) ? 0 : oldTab.length;int oldThr = threshold;int newCap, newThr = 0 ;if (oldCap > 0 ) { if (oldCap >= MAXIMUM_CAPACITY) { threshold = Integer.MAX_VALUE; return oldTab; } else if ((newCap = oldCap << 1 ) < MAXIMUM_CAPACITY && oldCap >= DEFAULT_INITIAL_CAPACITY) newThr = oldThr << 1 ; } else if (oldThr > 0 ) newCap = oldThr; else { newCap = DEFAULT_INITIAL_CAPACITY; newThr = (int )(DEFAULT_LOAD_FACTOR * DEFAULT_INITIAL_CAPACITY); } if (newThr == 0 ) { float ft = (float )newCap * loadFactor; newThr = (newCap < MAXIMUM_CAPACITY && ft < (float )MAXIMUM_CAPACITY ? (int )ft : Integer.MAX_VALUE); } threshold = newThr;
get(Object key) get方法就更好理解了,首先还是通过hash值找到数组下标,然后通过数组下标获取的实际的元素。然后判断一下当前节点key的hash值是否与第一个节点相同,相同则直接返回结果。
如果不同,这个时候,就得看第一个节点后的下一个节点是采用的红黑树还是使用的链表。然后再根据key的hash去取值即可。
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 final Node<K,V> getNode (int hash, Object key) { Node<K,V>[] tab; Node<K,V> first, e; int n; K k; if ((tab = table) != null && (n = tab.length) > 0 && (first = tab[(n - 1 ) & hash]) != null ) { if (first.hash == hash && ((k = first.key) == key || (key != null && key.equals(k)))) return first; if ((e = first.next) != null ) { if (first instanceof TreeNode) return ((TreeNode<K,V>)first).getTreeNode(hash, key); do { if (e.hash == hash && ((k = e.key) == key || (key != null && key.equals(k)))) return e; } while ((e = e.next) != null ); } } return null ; }