HashMap
Posted mdc1771344
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HashMap
- 结构: 数组+链表(红黑树)
放数据:
public V put(K key, V value) { return putVal(hash(key), key, value, false, true); }
key的hash(): key==null,其hash为0,不为null其值为key的高16位异或key的低16位(h右移位16的值),这样会把这个hashcode的位的用上,因为接下来通过hash值判断数组的位置时使用的是01111这样的与运算,如果直接使用hashcode会使得hashcode的高位没有意义
hashcode与equals的关系: equals相等则hashcode一定相等,反之不一定,因为两个key相同的键值对数据就是hashcode相同但是equals不同
static final int hash(Object key) { int h; return (key == null) ? 0 : (h = key.hashCode()) ^ (h >>> 16); }
链表节点:
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; }
数组大小: 默认16,最大值1 << 30
static final int DEFAULT_INITIAL_CAPACITY = 1 << 4; static final int MAXIMUM_CAPACITY = 1 << 30;
数组扩容: 条件: 当前数组的容量被使用了0.75倍之后进行2倍扩容
static final float DEFAULT_LOAD_FACTOR = 0.75f;
链表转变: 当链表节点增加大于等于8时转换成红黑树,减少到6时有红黑树转换成普通链表
static final int TREEIFY_THRESHOLD = 8; static final int UNTREEIFY_THRESHOLD = 6;
putVal
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;//通过resize()初始化或者加倍数组大小 //节点为空 if ((p = tab[i = (n - 1) & hash]) == null)//用01111与hash进行与运算(计算机执行比较快,目的与取模一样) tab[i] = newNode(hash, key, value, null);//当该数组位置没有元素,直接把Node节点赋给该数组位置 //节点非空 else { Node<K,V> e; K k; //添加元素的key == p.key if (p.hash == hash &&((k = p.key) == key || (key != null && key.equals(k)))) e = p; //添加元素的key != p.key,放入红黑树 else if (p instanceof TreeNode) e = ((TreeNode<K,V>)p).putTreeVal(this, tab, hash, key, value); //添加元素的key != p.key,放入普通链表 else { for (int binCount = 0; ; ++binCount) { if ((e = p.next) == null) { p.next = newNode(hash, key, value, null); if (binCount >= TREEIFY_THRESHOLD - 1) // 节点大于等于8时由链表转换为红黑树 treeifyBin(tab, hash); break; } if (e.hash == hash && ((k = e.key) == key || (key != null && key.equals(k))))//遍历链表时,存在相同key节点 break; p = e; } } //上面的for循环如果发现相同的key的节点就把e置为改节点,否则置为null 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)//扩容: size为数组所使用的大小,threshold为0.75*数组容量 resize(); afterNodeInsertion(evict); return null;
为什么数组两倍扩容? ---->因为需要使用数组长度(n)-1进行与计算,需要生成后位都是1的数,初始容量为16(10000),两倍扩容后就可以产生这个效果(例:10000-1=01111,100000000-1=011111111)
初始化或者加倍数组大小
final Node<K,V>[] resize() { Node<K,V>[] oldTab = table; int oldCap = (oldTab == null) ? 0 : oldTab.length; int oldThr = threshold; int newCap, newThr = 0; if (oldCap > 0) { if (oldCap >= MAXIMUM_CAPACITY) { //已经达到最大数组,不扩容了 threshold = Integer.MAX_VALUE; return oldTab; } else if ((newCap = oldCap << 1) < MAXIMUM_CAPACITY &&oldCap >= DEFAULT_INITIAL_CAPACITY) newThr = oldThr << 1; // double threshold } else if (oldThr > 0) // initial capacity was placed in threshold newCap = oldThr; else { // zero initial threshold signifies using defaults 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; @SuppressWarnings({"rawtypes","unchecked"}) Node<K,V>[] newTab = (Node<K,V>[])new Node[newCap]; table = newTab; if (oldTab != null) { //遍历数组 for (int j = 0; j < oldCap; ++j) { Node<K,V> e; if ((e = oldTab[j]) != null) { oldTab[j] = null; if (e.next == null) newTab[e.hash & (newCap - 1)] = e; //拆分红黑树 else if (e instanceof TreeNode) ((TreeNode<K,V>)e).split(this, newTab, j, oldCap); //拆分链表 else { // preserve order Node<K,V> loHead = null, loTail = null; Node<K,V> hiHead = null, hiTail = null; Node<K,V> next; do { next = e.next; if ((e.hash & oldCap) == 0) {//链表节点迁移只有两种可能,原地不动和往后移oldCap if (loTail == null) loHead = e; else loTail.next = e; loTail = e; } else { if (hiTail == null) hiHead = e; else hiTail.next = e; hiTail = e; } } while ((e = next) != null); if (loTail != null) { loTail.next = null; newTab[j] = loHead; } if (hiTail != null) { hiTail.next = null; newTab[j + oldCap] = hiHead; } } } } } return newTab; }
数组扩容后节点迁移
扩容前: 0x0101 & 01111 = 0101 扩容后: 0x0101 & 11111 = x0101
x只能为0或1,当为0时节点保持不动,当为1时,节点位置增加了10000也就是oldCap(扩容前数组的容量)
设置初始容量
initialCapacity = (存储元素个数 / 负载因子) + 1
null值(<<阿里巴巴Java开发手册>>)
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