HashMap的个别方法

 1 扩容的方法
 2 final Node<K,V>[] resize() {
 3     Node<K,V>[] oldTab = table; 
 4     int oldCap = (oldTab == null) ? 0 : oldTab.length; //取得table的长度
 5     int oldThr = threshold; //取得扩容的阈值
 6     int newCap, newThr = 0; 
 7     if (oldCap > 0) { //如果table长度大于0
 8         if (oldCap >= MAXIMUM_CAPACITY) { //判断是否大于最大容量
 9             threshold = Integer.MAX_VALUE;  //把 2的31次方赋值给阈值 ，也就是说这辈子不能扩容了
10             return oldTab; //直接返回
11         } 
12         else if ((newCap = oldCap << 1) < MAXIMUM_CAPACITY &&
13                  oldCap >= DEFAULT_INITIAL_CAPACITY)
14                  //如果 table 的长度乘以2 小于最大容量 并且  table的长度大于等于默认初始容量
15             newThr = oldThr << 1; // double threshold
16     }
17     else if (oldThr > 0) // initial capacity was placed in threshold
18         newCap = oldThr;  //如果原来的threshold大于0 ， 则把此当作新的容量
19     else {               // zero initial threshold signifies using defaults
20         newCap = DEFAULT_INITIAL_CAPACITY; //如果threshold也是0 ，则给默认的容量 16
21         newThr = (int)(DEFAULT_LOAD_FACTOR * DEFAULT_INITIAL_CAPACITY);
22         //新的阈值为 默认的负载因子*默认的初始容量
23     }
24     if (newThr == 0) { 
25         float ft = (float)newCap * loadFactor;//第二步else if并没有计算 新的阈值
26         newThr = (newCap < MAXIMUM_CAPACITY && ft < (float)MAXIMUM_CAPACITY ?
27                   (int)ft : Integer.MAX_VALUE);
28                   //如果 新的容量小于最大容量 并且 新的阈值小于最大容量 ？新的阈值：2的31次方
29     }
30     threshold = newThr; 
31     @SuppressWarnings({"rawtypes","unchecked"})
32         Node<K,V>[] newTab = (Node<K,V>[])new Node[newCap];
33     table = newTab; //扩容完毕
34     if (oldTab != null) {  //把原来table中的值迁移到新的table中
35         for (int j = 0; j < oldCap; ++j) { 
36             Node<K,V> e;
37             if ((e = oldTab[j]) != null) {
38                 oldTab[j] = null;
39                 if (e.next == null)
40                     newTab[e.hash & (newCap - 1)] = e;
41                 else if (e instanceof TreeNode)
42                     ((TreeNode<K,V>)e).split(this, newTab, j, oldCap);
43                 else { // preserve order
44                     Node<K,V> loHead = null, loTail = null;
45                     Node<K,V> hiHead = null, hiTail = null;
46                     Node<K,V> next;
47                     do {
48                         next = e.next;
49                         if ((e.hash & oldCap) == 0) {
50                             if (loTail == null)
51                                 loHead = e;
52                             else
53                                 loTail.next = e;
54                             loTail = e;
55                         }
56                         else {
57                             if (hiTail == null)
58                                 hiHead = e;
59                             else
60                                 hiTail.next = e;
61                             hiTail = e;
62                         }
63                     } while ((e = next) != null);
64                     if (loTail != null) {
65                         loTail.next = null;
66                         newTab[j] = loHead;
67                     }
68                     if (hiTail != null) {
69                         hiTail.next = null;
70                         newTab[j + oldCap] = hiHead;
71                     }
72                 }
73             }
74         }
75     }
76     return newTab;//返回
77 }

Put方法
public V put(K key, V value) {
    return putVal(hash(key), key, value, false, true);
}

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) //如果table为null 或者 长度为0
        n = (tab = resize()).length; //  调用resize生成数组table, 并令tab指向数组table
    if ((p = tab[i = (n - 1) & hash]) == null) //通过表的长度与hash进行位运算
                        ，取得一个合适下标，如果该下标当前为null，创建新的节点
        tab[i] = newNode(hash, key, value, null);
    else { //如果散列冲突
        Node<K,V> e; K k;
        if (p.hash == hash &&  //如果该节点上这一对的 hash 与当前要放入的这一对 hash相同，且key相同
            ((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) { //如果p节点的下一个为null ，p.next指向当前要放入的Node
                    p.next = newNode(hash, key, value, null);
                    if (binCount >= TREEIFY_THRESHOLD - 1) // -1 for 1st
                        treeifyBin(tab, hash); //  0.1.2.3.4.5.6.7 循环了八次，放入当前节点，即
                                                                    超过八个节点，则转为红黑树
                    break; //over 退出循环
                }
                if (e.hash == hash &&
                    ((k = e.key) == key || (key != null && key.equals(k))))
                    break; // 如果后面节点中由key相同 over 退出循环
                p = e;   
            }
        }
        if (e != null) { // existing mapping for key
            V oldValue = e.value;
            if (!onlyIfAbsent || oldValue == null)// onlyIfAbsent为true表示仅当<key,value>不存在时进行插入, 为false表示强制覆盖;
                e.value = value; 
            afterNodeAccess(e);
            return oldValue; 
        }
    }
    ++modCount;
    if (++size > threshold)  //判断是否需要扩容
        resize(); 
    afterNodeInsertion(evict);
    return null;

remove方法
public V remove(Object key) {
    Node<K,V> e;
    return (e = removeNode(hash(key), key, null, false, true)) == null ?
        null : e.value;
}

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 && //如果table不为空，并且长度大于0并且 
                        //通过key的hash值与长度所计算出来的数组下标所对应的引用不为空
        (p = tab[index = (n - 1) & hash]) != null) {
        Node<K,V> node = null, e; K k; V v;
        if (p.hash == hash && //如果hash值相同，且key也相同
            ((k = p.key) == key || (key != null && key.equals(k))))
            node = p; //把这个节点拿出来
        else if ((e = p.next) != null) {//如果hash相同 ，key不相同，并且下一个不是null
            if (p instanceof TreeNode)
                node = ((TreeNode<K,V>)p).getTreeNode(hash, key);
            else {
                do { //找到hash相同 key相同的节点
                    if (e.hash == hash &&
                        ((k = e.key) == key ||
                         (key != null && key.equals(k)))) {
                        node = e; //找到了
                        break;
                    }
                    p = e;
                } while ((e = e.next) != null); //循环整个链表，知道为空 此时 p.next = e
            }
        }
        //matchValue if true only remove if value is equal
        //如果为真，则仅在值相等时删除 ！false
        if (node != null && (!matchValue || (v = node.value) == value ||
                             (value != null && value.equals(v)))) {
            if (node instanceof TreeNode)
                ((TreeNode<K,V>)node).removeTreeNode(this, tab, movable);
            else if (node == p) 
                tab[index] = node.next;
            else
                p.next = node.next; //直接把p.next 指向 当前要删除的节点的next 即跳过了当前节点
            ++modCount;
            --size;
            afterNodeRemoval(node);
            return node; 
        }
    }
    return null;//没找着
}

链表转红黑树
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(); //如果table的长度小于 MIN_TREEIFY_CAPACITY  即64 则会选择扩容
    else if ((e = tab[index = (n - 1) & hash]) != null) {
        TreeNode<K,V> hd = null, tl = null;
        do {
            TreeNode<K,V> p = replacementTreeNode(e, null);
            if (tl == null)
                hd = p;
            else {
                p.prev = tl;
                tl.next = p;
            }
            tl = p;
        } while ((e = e.next) != null);
        if ((tab[index] = hd) != null)
            hd.treeify(tab);
    }
}