Map--HashMap实现分析

HashMap:

利用数组和链表的数据结构存储<key,value>

链表：

static class Node<K,V> implements Map.Entry<K,V>数组：
transient Node<K,V>[] table;储存过程：
1对key进行哈希

2根据哈希值找到数组中对应位置

3如果该位置没有链表则作为头部放入，如果该位置存在链表则遍历链表，如果没找到则在尾部加入<key,value>
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);//没有链表作为头部放入
else {
Node<K,V> e; K k;
if (p.hash == hash &&
((k = p.key) == key || (key != null && key.equals(k))))
e = p;//更新指针指向表头
else if (p instanceof TreeNode)
e = ((TreeNode<K,V>)p).putTreeVal(this, tab, hash, key, value);//树结构在这里不讨论
else {
for (int binCount = 0; ; ++binCount) {
if ((e = p.next) == null) {
p.next = newNode(hash, key, value, null);//链表中没找到在尾部新增<key,value>
if (binCount >= TREEIFY_THRESHOLD - 1) // -1 for 1st//链表长度再分配在这里不讨论
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//更新<key,value>
V oldValue = e.value;
if (!onlyIfAbsent || oldValue == null)
e.value = value;
afterNodeAccess(e);
return oldValue;
}
}
++modCount;
if (++size > threshold)
resize();
afterNodeInsertion(evict);
return null;
}读取过程：
1对key进行哈希

2根据哈希值找到数组中对应位置的头部节点

3判断key是否与头部以及链表其他节点相同

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 && // always check first node
((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;
}

resize：
HashMap设置了三个参数：

threshold当前允许元素最大数

loadFactor负载因子

capacity数组大小

loadFactor=threshold/capacity

如果loadFactor太小则浪费空间，太大在每次put和get的时候线性时间会比较长，因此一般固定loadFactor，默认为0.75，当元素数量达到threshold则进行resize扩大capacity以及threshold

resize步骤：

1如果table为空采用默认的capacity和threshold，如果不为空则扩大一倍

2将原来table中的<key,value>放入新的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;
@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;//放入新的table
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;
}hash技巧：
由于数组长度有限，如果仅仅利用hashcode和数组长度n-1与操作，那么高位不同低位相同将会哈希到同一个桶中，增加了碰撞率。因此采用高位计算，将hashcode右移把高位和低位进行异或。

<
4000
pre class="html">static final int hash(Object key) {
int h;
return (key == null) ? 0 : (h = key.hashCode()) ^ (h >>> 16);
}同时在计算数组中的对应位置时，与数组长度n-1进行与操作，而数组长度都是2的指数次，这样使n-1的二进制码为全1，就可以避免0到n-1之间不同的值与操作完被分配到相同的位置。下面的函数保证了数组长度为2的指数次：
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;
}