1.HashMap 的原理,内部数据结构?
底层使用哈希表(数组加链表),当链表过长会将链表转成 红黑树 以实现0(logn)时间复杂度内查找
2.将下HashMap中put方法过程?
i.对key求hash值,然后再计算 下标
ii:如果没有碰撞,直接放入桶中
iii:如果碰撞了,以链表的方式连接到后面
iv:如果链表的长度超过阀值(TREEIFY_THRESHOLD==8),就把链表转成红黑树。
v:如果节点已经存在就替换旧值
vi:如果桶满了(容量*加载引子),就需要resize。
3.HashMap中hash函数怎么是实现?还有那些hash的实现方式?
i:高16bit不变,低16bit与高16bit做一个与或
ii: (n-1)&hash-->得到下标
iii:还有那些hash的实现方式:可以参考之前的博客 Effective java 学习笔记--hashcode()
4.hashmap如何解决冲突,讲一下扩容过程,加入一个值在原数组中,现在移动之后变为新数组,位置肯定改变了,那是什么定位到这个值新数组中的位置
。将新节点加到链表后
。容量扩充为原来的两倍,然后对每个节点重新计算hash值
。这个值只可能在两个地方,一个是原下标的地方,另一个是在下标为<原下标+原容量>的位置
5 抛开hashMap, hash冲突有哪些解决办法?
开放地址,链地址法
6.针对hashmap中的某个entry链太长,查找的时间复杂度可能达到O(n),怎么优化?
。 将链表转化为红黑树,JDK1.8 已经实现
数据在计算机中的存储结构(数据结构)
1.数组
2.链表
23树结构
数组(ArrayList)
链表(linklist)
双向链表 :查询效率低 增删效率高
HashMap 数组加链表
数组的表示
数组的初始化容量 最大容量 加载因子
/** * The default initial capacity - MUST be a power of two. */ static final int DEFAULT_INITIAL_CAPACITY = 1 << 4; // aka 16 /** * The maximum capacity, used if a higher value is implicitly specified * by either of the constructors with arguments. * MUST be a power of two <= 1<<30. */ static final int MAXIMUM_CAPACITY = 1 << 30; /** * The load factor used when none specified in constructor. */ static final float DEFAULT_LOAD_FACTOR = 0.75f;
数组的大小可能不够用 :扩大
什么时候扩大: 16*0.72=12(原大小*加载因子) 数组扩大的标准 首次用到12的时候扩大
链表的长度可以无限变大吗?不可以 链表的长度超过treeify_threshold=8时 链表转化为红黑树
/** * The bin count threshold for using a tree rather than list for a * bin. Bins are converted to trees when adding an element to a * bin with at least this many nodes. The value must be greater * than 2 and should be at least 8 to mesh with assumptions in * tree removal about conversion back to plain bins upon * shrinkage. */ static final int TREEIFY_THRESHOLD = 8;
链表和红黑树有各自最优效率的元素数量 当红黑树中的元素小于 untreeify_threshold=6时将转化回链表
/** * The bin count threshold for untreeifying a (split) bin during a * resize operation. Should be less than TREEIFY_THRESHOLD, and at * most 6 to mesh with shrinkage detection under removal. */ static final int UNTREEIFY_THRESHOLD = 6;
每次向数组中添加元素 都会记录已经使用的格子数量,当size大于 16*0.72=12(原大小*加载因子 数组扩大的标准)时 resize()
来了一个key,value组成了Node节点之后,这个节点到底该何去何从?
生产出一个算法 hash()算法
1, 返回一个 int类型的值
key,value ------>key(Object) 有个hashcode()方法得到key的hash值 :3254239
2, 0~15之间 数组大小的范围内
hash%16=0~15
3,尽可能利用数组的每一个位置
注意node节点的属性 key的hash, key,value,next下一个节点
/** * Basic hash bin node, used for most entries. (See below for * TreeNode subclass, and in LinkedHashMap for its Entry subclass.) */ 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; } }
新来一个节点调用put() 根据hash(key)计算hash值(对hashCode()的值进行高十六位和低十六位运算)
在putval()中新节点的到来分为三种情况
1 table数组原来的位置为空
2 table数组原来的位置不为空,且下面是链表的结构
3 table数组原来的位置不为空,且下面是红黑树的结构
/** * Associates the specified value with the specified key in this map. * If the map previously contained a mapping for the key, the old * value is replaced. * * @param key key with which the specified value is to be associated * @param value value to be associated with the specified key * @return the previous value associated with <tt>key</tt>, or * <tt>null</tt> if there was no mapping for <tt>key</tt>. * (A <tt>null</tt> return can also indicate that the map * previously associated <tt>null</tt> with <tt>key</tt>.) */ public V put(K key, V value) { return putVal(hash(key), key, value, false, true); }
//取key的hashcode() 与 hashcode()的低十六位运算结果 做异或运算
static final int hash(Object key) { int h; return (key == null) ? 0 : (h = key.hashCode()) ^ (h >>> 16); }
/** * 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)//根据key的hash值计算该note节点在数组中的角标 判断该角标位数元素是否为空 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))//table数组原来的位置不为空,且下面是红黑树的结构 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// 链表的元素超过八个装换为红黑树 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) //超过12 则去扩容 resize(); afterNodeInsertion(evict); return null; }
/** * Initializes or doubles table size. 初始化或扩容table数组的size If null, allocates in * accord with initial capacity target held in field threshold. * Otherwise, because we are using power-of-two expansion, the * elements from each bin must either stay at same index, or move * with a power of two offset in the new table. * * @return the table */ 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) {//是否大于最大值 threshold = Integer.MAX_VALUE; return oldTab; } else if ((newCap = oldCap << 1) < MAXIMUM_CAPACITY && oldCap >= DEFAULT_INITIAL_CAPACITY) newThr = oldThr << 1; // double threshold扩容两倍 } else if (oldThr > 0) // initial capacity was placed in threshold newCap = oldThr; else { // zero initial threshold signifies using defaults newCap = DEFAULT_INITIAL_CAPACITY;//默认大小16 newThr = (int)(DEFAULT_LOAD_FACTOR * DEFAULT_INITIAL_CAPACITY);//需扩容时的容量 16*0.75=12 } 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];//初始化一个默认大小16的node数组 table = newTab; if (oldTab != null) { 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; else if (e instanceof TreeNode) ((TreeNode<K,V>)e).split(this, newTab, j, oldCap); 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) { 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; }
分析
(n - 1) & hash
hash为 根据hash(key)计算hash的值(对hashCode()的值与hashCode()的低16位进行异或运算) 例 :3254239
n为数组初始化的长度16
即为 15&3254239 转化为 2进制 &运算 结果在0000到1111(0~15)之间
所以数组的大小永远是2的n次幂 以保证得到的结果在0到15之间
如何保证尽可能利用数组的每一个位置?
需保证hash的值足够分散 对对hashCode()的值与hashCode()的低16位进行异或运算
static final int hash(Object key) { int h; return (key == null) ? 0 : (h = key.hashCode()) ^ (h >>> 16); }