1.数据结构分析
(1)HashMap分析 : https://www.cnblogs.com/kubixuesheng/p/6144535.html put、get、resize分析
jdk 1.7 数组+链表
jdk 1.8 数组+链表(红黑树)
红黑树与avl树区别：
avl树需要完全平衡，因此插入、修改都要调整；查找更快，但是插入和修改效率较差；红黑树取了一个中间平衡
红黑树定义：
性质1. 节点是红色或黑色。
性质2. 根节点是黑色。
性质3 每个红色节点的两个子节点都是黑色。(从每个叶子到根的所有路径上不能有两个连续的红色节点)
性质4. 从任一节点到其每个叶子的所有路径都包含相同数目的黑色节点。
https://baijiahao.baidu.com/s?id=1618550070727689060&wfr=spider&for=pc
2.源码篇
HashMap https://blog.csdn.net/ns_code/article/details/36034955 key,value 都允许为null
public V put(K key, V value) { return putVal(hash(key), key, value, false, true); } /** * Implements Map.put and related methods. * * @param hash hash for key * @param key the key * @param value the value to put * @param onlyIfAbsent if true, don't change existing value * @param evict if false, the table is in creation mode. * @return previous value, or null if none */ 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; }
HashTable https://blog.csdn.net/ns_code/article/details/36191279 value不允许为空
public synchronized V put(K key, V value) { // Make sure the value is not null if (value == null) { throw new NullPointerException(); } // Makes sure the key is not already in the hashtable. Entry<?,?> tab[] = table; int hash = key.hashCode(); int index = (hash & 0x7FFFFFFF) % tab.length; @SuppressWarnings("unchecked") Entry<K,V> entry = (Entry<K,V>)tab[index]; for(; entry != null ; entry = entry.next) { if ((entry.hash == hash) && entry.key.equals(key)) { V old = entry.value; entry.value = value; return old; } } addEntry(hash, key, value, index); return null; }
CurrentHashMap https://www.jianshu.com/p/865c813f2726 key,value都不允许为空
public V put(K key, V value) { return putVal(key, value, false); } /** Implementation for put and putIfAbsent */ final V putVal(K key, V value, boolean onlyIfAbsent) { if (key == null || value == null) throw new NullPointerException(); int hash = spread(key.hashCode()); int binCount = 0; for (Node<K,V>[] tab = table;;) { Node<K,V> f; int n, i, fh; if (tab == null || (n = tab.length) == 0) tab = initTable(); else if ((f = tabAt(tab, i = (n - 1) & hash)) == null) { if (casTabAt(tab, i, null, new Node<K,V>(hash, key, value, null))) break; // no lock when adding to empty bin } else if ((fh = f.hash) == MOVED) tab = helpTransfer(tab, f); else { V oldVal = null; synchronized (f) { if (tabAt(tab, i) == f) { if (fh >= 0) { binCount = 1; for (Node<K,V> e = f;; ++binCount) { K ek; if (e.hash == hash && ((ek = e.key) == key || (ek != null && key.equals(ek)))) { oldVal = e.val; if (!onlyIfAbsent) e.val = value; break; } Node<K,V> pred = e; if ((e = e.next) == null) { pred.next = new Node<K,V>(hash, key, value, null); break; } } } else if (f instanceof TreeBin) { Node<K,V> p; binCount = 2; if ((p = ((TreeBin<K,V>)f).putTreeVal(hash, key, value)) != null) { oldVal = p.val; if (!onlyIfAbsent) p.val = value; } } } } if (binCount != 0) { if (binCount >= TREEIFY_THRESHOLD) treeifyBin(tab, i); if (oldVal != null) return oldVal; break; } } } addCount(1L, binCount); return null; }
