我们接着看
remove 实现
/**
* Removes the mapping for the specified key from this map if present.
*
* @param key key whose mapping is to be removed from the map
* @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 remove(Object key) {
Node<K,V> e;
return (e = removeNode(hash(key), key, null, false, true)) == null ? null : e.value;
}
/**
* Implements Map.remove and related methods
*
* @param hash hash for key
* @param key the key
* @param value the value to match if matchValue, else ignored
* @param matchValue if true only remove if value is equal
* @param movable if false do not move other nodes while removing
* @return the node, or null if none
*/
final Node<K,V> removeNode(int hash, Object key, Object value,boolean matchValue, boolean movable) {
Node<K,V>[] tab; Node<K,V> p; int n, index;
if ((tab = table) != null && (n = tab.length) > 0 && (p = tab[index = (n - 1) & hash]) != null) {
// 找到hash 对应的 数组元素
Node<K,V> node = null, e; K k; V v;
if (p.hash == hash && ((k = p.key) == key || (key != null && key.equals(k))))
node = p;
else if ((e = p.next) != null) {
if (p instanceof TreeNode)
// 链表有 元素，如果是红黑树结构
node = ((TreeNode<K,V>)p).getTreeNode(hash, key);
else {
// 链表有 元素，如果是链表结构
do {
if (e.hash == hash &&
((k = e.key) == key ||
(key != null && key.equals(k)))) {
node = e;
break;
}
p = e;
} while ((e = e.next) != null);
}
}
if (node != null && (!matchValue || (v = node.value) == value || (value != null && value.equals(v)))) {
// 如果找到元素，则开始 进行移除元素操作
if (node instanceof TreeNode)
// 在红黑树上，移除节点
((TreeNode<K,V>)node).removeTreeNode(this, tab, movable);
else if (node == p)
// 在数组上，移除节点
tab[index] = node.next;
else
// 在链表上，移除节点
p.next = node.next;
++modCount;
--size;
afterNodeRemoval(node);
return node;
}
}
return null;
}
可以看到不同的存储结构他采用了不同的查找方式，以及移除方式，我们来看看他的查找方式
getTreeNode实现
/**
* Calls find for root node.
*/
final TreeNode<K,V> getTreeNode(int h, Object k) {
return ((parent != null) ? root() : this).find(h, k, null);
}
/**
* Finds the node starting at root p with the given hash and key.
* The kc argument caches comparableClassFor(key) upon first use
* comparing keys.
*/
final TreeNode<K,V> find(int h, Object k, Class<?> kc) {
TreeNode<K,V> p = this;
do {
int ph, dir; K pk;
TreeNode<K,V> pl = p.left, pr = p.right, q;
if ((ph = p.hash) > h)
p = pl;
else if (ph < h)
p = pr;
else if ((pk = p.key) == k || (k != null && k.equals(pk)))
return p;
else if (pl == null)
p = pr;
else if (pr == null)
p = pl;
else if ((kc != null ||
(kc = comparableClassFor(k)) != null) &&
(dir = compareComparables(kc, k, pk)) != 0)
p = (dir < 0) ? pl : pr;
else if ((q = pr.find(h, k, kc)) != null)
return q;
else
p = pl;
} while (p != null);
return null;
}
树形结构特点就是查找非常快速，使用了二分查找，对比hash 与 左右分支的节点hash来确认查找的结果，以及接下来去哪个分支上查找。删除的实现，就比较复杂了，当链表的节点数大于8的时候，链表会转化成红黑树，那是不是红黑树的节点数 小于等 8，会还原成 链表呢，带着这个问题，来看看删除的实现：
removeTreeNode实现
/**
* Removes the given node, that must be present before this call.
* This is messier than typical red-black deletion code because we
* cannot swap the contents of an interior node with a leaf
* successor that is pinned by "next" pointers that are accessible
* independently during traversal. So instead we swap the tree
* linkages. If the current tree appears to have too few nodes,
* the bin is converted back to a plain bin. (The test triggers
* somewhere between 2 and 6 nodes, depending on tree structure).
*/
final void removeTreeNode(HashMap<K,V> map, Node<K,V>[] tab,boolean movable) {
int n;
if (tab == null || (n = tab.length) == 0)
return;
int index = (n - 1) & hash;
TreeNode<K,V> first = (TreeNode<K,V>)tab[index], root = first, rl;
TreeNode<K,V> succ = (TreeNode<K,V>)next, pred = prev;
if (pred == null)
tab[index] = first = succ;
else
pred.next = succ;
if (succ != null)
succ.prev = pred;
if (first == null)
return;
if (root.parent != null)
root = root.root();
if (root == null || root.right == null ||
(rl = root.left) == null || rl.left == null) {
// 果然当红黑树的结构太小，就会转化成链表结构
// 但是转换条件并不是 节点个数的问题，而是树形结构，
// 这里的判断条件在于红黑树的层级，少于3层树形结构，就会被转化成链表的存储结构
tab[index] = first.untreeify(map); // too small
return;
}
// 接下来就是 红黑树删除节点算法了，到时候我们花一篇文章的时间来介绍红黑树结构
// 以及他的操作算法节点说明
TreeNode<K,V> p = this, pl = left, pr = right, replacement;
if (pl != null && pr != null) {
TreeNode<K,V> s = pr, sl;
while ((sl = s.left) != null) // find successor
s = sl;
boolean c = s.red; s.red = p.red; p.red = c; // swap colors
TreeNode<K,V> sr = s.right;
TreeNode<K,V> pp = p.parent;
if (s == pr) { // p was s's direct parent
p.parent = s;
s.right = p;
}
else {
TreeNode<K,V> sp = s.parent;
if ((p.parent = sp) != null) {
if (s == sp.left)
sp.left = p;
else
sp.right = p;
}
if ((s.right = pr) != null)
pr.parent = s;
}
p.left = null;
if ((p.right = sr) != null)
sr.parent = p;
if ((s.left = pl) != null)
pl.parent = s;
if ((s.parent = pp) == null)
root = s;
else if (p == pp.left)
pp.left = s;
else
pp.right = s;
if (sr != null)
replacement = sr;
else
replacement = p;
}
else if (pl != null)
replacement = pl;
else if (pr != null)
replacement = pr;
else
replacement = p;
if (replacement != p) {
TreeNode<K,V> pp = replacement.parent = p.parent;
if (pp == null)
root = replacement;
else if (p == pp.left)
pp.left = replacement;
else
pp.right = replacement;
p.left = p.right = p.parent = null;
}
TreeNode<K,V> r = p.red ? root : balanceDeletion(root, replacement);
if (replacement == p) { // detach
TreeNode<K,V> pp = p.parent;
p.parent = null;
if (pp != null) {
if (p == pp.left)
pp.left = null;
else if (p == pp.right)
pp.right = null;
}
}
if (movable)
moveRootToFront(tab, r);
}
接下来，我们来分析一下他的扩容操作，扩容大小是多少，以及扩容的时机是什么时候等等
reSize()实现
/**
* 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;
// 存储数组的 长度
int oldCap = (oldTab == null) ? 0 : oldTab.length;
// 存储 整个HashMap的 存储数据量
int oldThr = threshold;
int newCap, newThr = 0;
if (oldCap > 0) {
if (oldCap >= MAXIMUM_CAPACITY) {
threshold = Integer.MAX_VALUE;
return oldTab;
}
// 如果老的数组长度 不为零，且小于最大数组长度大小 MAXIMUM_CAPACITY
// 新的数组长度是原来的2倍大小
else if ((newCap = oldCap << 1) < MAXIMUM_CAPACITY &&
oldCap >= DEFAULT_INITIAL_CAPACITY)、
// 如果新的数组长度 小于最大数组长度大小 MAXIMUM_CAPACITY
// 如果满足要求，则HashMap容量变成原来的2倍
newThr = oldThr << 1; // double threshold
}
else if (oldThr > 0) // initial capacity was placed in threshold
newCap = oldThr;
else { // zero initial threshold signifies using defaults
// 如果未初始化，则赋值 初始化数据
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;
// 以下就是 在容量发生变化的时候，存储数据进行重新排列赋值操作
if (oldTab != null) {
for (int j = 0; j < oldCap; ++j) {
Node<K,V> e;
if ((e = oldTab[j]) != null) {
// 老的数组数据被赋值成空
oldTab[j] = null;
if (e.next == null)
// 如果当前的桶 中只有一个元素，则直接赋值给新的 数组中
newTab[e.hash & (newCap - 1)] = e;
else if (e instanceof TreeNode)
// 如果是红黑树，则使用分离的算法，
((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;
// 因为 扩容大小是 原有的大小的两倍，使用 e.hash & oldCap 算法计算
// 计算出为 0 代表当前节点的hash二进制数 与 oldCap的二进制数 不存在相同位为 1的，如果扩容之后重新计算 的话，index 还是会为 j
// 计算出不为 0 代表当前节点的hash 二进制数 与 oldCap 的二进制数 存在相同位 为 1的，如果扩容之后重新计算的话，index 会变成 j + oldCap
// 如果为 0 ，则存储在新的数组中的位置 还是为 j
// 如果不为 0 ，则 存储位置在新的数组中的位置 为 j + oldCap
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;
}
bin : 箱子，也可以称作为 桶，对应的就是数组中 index 对应的存储结构了，可以能是链式，可能是红黑树
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