1 put过程
/**
* 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;
}
1.判断散列表是否为null,如果为null,重新初始化散列表
2.如果不为null并且没有发生碰撞,直接添加元素到散列表中
3.如果发生碰撞了,并且要插入的元素的桶的hash和key都相等,记录下来,将新值覆盖旧值
4.如果桶hash与key不想等,并且该节点是红黑树结构,调用树的插入方法
5.如果是链表结构,找到了key映射节点,就记录这个节点,退出循环。如果没有找到,在链表尾部插入节点。
插入后如果发现临界值大于TREEIFY_THRESHOLD,转成红黑树
2 解决hash冲突的一般过程
- 开放地址法:此处不留爷,我去下一家
- 分离链表法:就像窗口打饭,来一个打饭的,跟在后面就行了,构成链表
3 hashMap扩容阈值为何是2的整数幂?
~~ 初始容量为何是16
扩容代码:/**
* Initializes or doubles table 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) {
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; //threshold:扩容阈值loadFactor(0.75)*length(16)
@SuppressWarnings({"rawtypes","unchecked"})
Node<K,V>[] newTab = (Node<K,V>[])new Node[newCap];
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;
}
容量计算方法:
public HashMap(int initialCapacity, float loadFactor) {
......
this.threshold = tableSizeFor(initialCapacity);
}
static final int tableSizeFor(int cap) {
int n = cap - 1;
n |= n >>> 1;
n |= n >>> 2;
n |= n >>> 4;
n |= n >>> 8;
n |= n >>> 16;
return (n < 0) ? 1 : (n >= MAXIMUM_CAPACITY) ? MAXIMUM_CAPACITY : n + 1;
}
总结:为什么这里一定要指定容量为2的n次方呢?
当length总是 2 的n次方时,h& (length-1)运算等价于对length取模,也就是h%length,但是&比%具有更高的效率。这样也能使hashcode的取模结果更为平均,尽量的减少冲突
下面我们讲解下JDK1.8做了哪些优化。经过观测可以发现,我们使用的是2次幂的扩展(指长度扩为原来2倍),所以,元素的位置要么是在原位置,要么是在原位置再移动2次幂的位置。看下图可以明白这句话的意思,n为table的长度,图(a)表示扩容前的key1和key2两种key确定索引位置的示例,图(b)表示扩容后key1和key2两种key确定索引位置的示例,其中hash1是key1对应的哈希与高位运算结果。
元素在重新计算hash之后,因为n变为2倍,那么n-1的mask范围在高位多1bit(红色),因此新的index就会发生这样的变化:
因此,我们在扩充HashMap的时候,不需要像JDK1.7的实现那样重新计算hash,只需要看看原来的hash值新增的那个bit是1还是0就好了,是0的话索引没变,是1的话索引变成“原索引+oldCap”
4 为什么在链表长度为 8 的时候转红黑树,为啥不能是 9 是 10?
* threshold of 0.75, although with a large variance because of
* resizing granularity. Ignoring variance, the expected
* occurrences of list size k are (exp(-0.5) * pow(0.5, k) /
* factorial(k)). The first values are:
*
* 0: 0.60653066
* 1: 0.30326533
* 2: 0.07581633
* 3: 0.01263606
* 4: 0.00157952
* 5: 0.00015795
* 6: 0.00001316
* 7: 0.00000094
* 8: 0.00000006
* more: less than 1 in ten million
总结:随机hash码下,哈希表中频率分布遵循泊松分布(单位(时间)内出现的次数),长度为8,出现的概率为千万分之6,概率极低
5 为什么要转红黑树?
转红黑树代码:
/**
* Replaces all linked nodes in bin at index for given hash unless
* table is too small, in which case resizes instead.
*/
final void treeifyBin(Node<K,V>[] tab, int hash) {
int n, index; Node<K,V> e;
if (tab == null || (n = tab.length) < MIN_TREEIFY_CAPACITY)
resize();
......
}
问题博客
总结:① 在获取数据的时候,链表复杂度为O(n),而链表复杂度为O(logN)
② 为何不直接使用红黑树,因为树节点的大小是链表节点大小的两倍,所以只有在容器中包含足够的节点保证使用才用它。显然尽管转为树使得查找的速度更快,但是在节点数比较小的时候,此时对于红黑树来说内存上的劣势会
6 JDK1.8做了那些改变?
- hash表由原来的“数组+链表”——>”数组+链表+红黑树”
- 链表增加节点“头插法”——>”尾插法”(避免死锁)
参考文档