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判断键值对数组table[i]是否为空或为null，否则执行resize()进行扩容；
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根据键值key计算hash值得到插入的数组索引i，如果table[i]==null，直接新建节点添加，转向6，如果table[i]不为空，转向3；
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判断table[i]的首个元素是否和key一样，如果相同直接覆盖value，否则转向4，这里的相同指的是hashCode以及equals；
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判断table[i] 是否为treeNode，即table[i] 是否是红黑树，如果是红黑树，则直接在树中插入键值对，否则转向5；
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遍历table[i]，判断链表长度是否大于8，大于8的话把链表转换为红黑树，在红黑树中执行插入操作，否则进行链表的插入操作；遍历过程中若发现key已经存在直接覆盖value即可；
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插入成功后，判断实际存在的键值对数量size是否超多了最大容量threshold，如果超过，进行扩容。
上源码
1 public V put(K key, V value) {
2 // 对key的hashCode()做hash
3 return putVal(hash(key), key, value, false, true);
4 }
5
6 final V putVal(int hash, K key, V value, boolean onlyIfAbsent,
7 boolean evict) {
8 Node<K,V>[] tab; Node<K,V> p; int n, i;
9 // 步骤①：tab为空则创建
10 if ((tab = table) == null || (n = tab.length) == 0)
11 n = (tab = resize()).length;
12 // 步骤②：计算index，并对null做处理
13 if ((p = tab[i = (n - 1) & hash]) == null)
14 tab[i] = newNode(hash, key, value, null);
15 else {
16 Node<K,V> e; K k;
17 // 步骤③：节点key存在，直接覆盖value
18 if (p.hash == hash &&
19 ((k = p.key) == key || (key != null && key.equals(k))))
20 e = p;
21 // 步骤④：判断该链为红黑树
22 else if (p instanceof TreeNode)
23 e = ((TreeNode<K,V>)p).putTreeVal(this, tab, hash, key, value);
24 // 步骤⑤：该链为链表
25 else {
26 for (int binCount = 0; ; ++binCount) {
27 if ((e = p.next) == null) {
28 p.next = newNode(hash, key,value,null);
//链表长度大于8转换为红黑树进行处理
29 if (binCount >= TREEIFY_THRESHOLD - 1) // -1 for 1st
30 treeifyBin(tab, hash);
31 break;
32 }
// key已经存在直接覆盖value
33 if (e.hash == hash &&
34 ((k = e.key) == key || (key != null && key.equals(k))))
35 break;
36 p = e;
37 }
38 }
39
40 if (e != null) { // existing mapping for key
41 V oldValue = e.value;
42 if (!onlyIfAbsent || oldValue == null)
43 e.value = value;
44 afterNodeAccess(e);
45 return oldValue;
46 }
47 }
48 ++modCount;
49 // 步骤⑥：超过最大容量 就扩容
50 if (++size > threshold)
51 resize();
52 afterNodeInsertion(evict);
53 return null;
54 }
二、HashMap之resize方法
final Node<K,V>[] resize() 大专栏 令人抓狂的HashMap - 长敏的小站class="o">{
Node<K,V>[] oldTab = table;
int oldCap = (oldTab == null) ? 0 : oldTab.length;
int oldThr = threshold;
int newCap, newThr = 0;
if (oldCap > 0) {
// 如果目前的容量已经达到最大容量，那就没必要扩容了
if (oldCap >= MAXIMUM_CAPACITY) {
threshold = Integer.MAX_VALUE;
return oldTab;
}
else if ((newCap = oldCap << 1) < MAXIMUM_CAPACITY &&
oldCap >= DEFAULT_INITIAL_CAPACITY)
newThr = oldThr << 1; // 容量threshold扩大一倍
}
else if (oldThr > 0) // 替换threshold
newCap = oldThr;
else { // 初始化
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;
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;
}
—未完待续……