HashMap夺命14问，你能坚持到第几问？

final Node[] resize() {        //oldTab：引用扩容前的哈希表        Node[] oldTab = table;        //oldCap：表示扩容前的table数组的长度        int oldCap = (oldTab == null) ? 0 : oldTab.length;        //获得旧哈希表的扩容阈值        int oldThr = threshold;        //newCap:扩容之后table数组大小        //newThr:扩容之后下次触发扩容的条件        int newCap, newThr = 0;        //条件成立说明hashMap中的散列表已经初始化过了，是一次正常扩容        if (oldCap > 0) {            //判断旧的容量是否大于等于最大容量，如果是，则无法扩容，并且设置扩容条件为int最大值，            //这种情况属于非常少数的情况            if (oldCap >= MAXIMUM_CAPACITY) {                threshold = Integer.MAX_VALUE;                return oldTab;            }//设置newCap新容量为oldCap旧容量的二倍（<<1）,并且<最大容量，而且>=16，则新阈值等于旧阈值的两倍            else if ((newCap = oldCap << 1) < MAXIMUM_CAPACITY &&                     oldCap >= DEFAULT_INITIAL_CAPACITY)                newThr = oldThr << 1; // double threshold        }        //如果oldCap=0并且边界值大于0，说明散列表是null，但此时oldThr>0        //说明此时hashMap的创建是通过指定的构造方法创建的,新容量直接等于阈值        //1.new HashMap(intitCap,loadFactor)        //2.new HashMap(initCap)        //3.new HashMap(map)        else if (oldThr > 0) // initial capacity was placed in threshold            newCap = oldThr;        //这种情况下oldThr=0;oldCap=0，说明没经过初始化，创建hashMap        //的时候是通过new HashMap()的方式创建的        else {               // zero initial threshold signifies using defaults            newCap = DEFAULT_INITIAL_CAPACITY;            newThr = (int)(DEFAULT_LOAD_FACTOR * DEFAULT_INITIAL_CAPACITY);        }        //newThr为0时，通过newCap和loadFactor计算出一个newThr        if (newThr == 0) {            //容量*0.75            float ft = (float)newCap * loadFactor;            newThr = (newCap < MAXIMUM_CAPACITY && ft < (float)MAXIMUM_CAPACITY ?                      (int)ft : Integer.MAX_VALUE);        }        threshold = newThr;        @SuppressWarnings({"rawtypes","unchecked"})                //根据上面计算出的结果创建一个更长更大的数组            Node[] newTab = (Node[])new Node[newCap];        //将table指向新创建的数组        table = newTab;        //本次扩容之前table不为null        if (oldTab != null) {            //对数组中的元素进行遍历            for (int j = 0; j < oldCap; ++j) {                //设置e为当前node节点                Node e;                //当前桶位数据不为空，但不能知道里面是单个元素，还是链表或红黑树，                //e = oldTab[j]，先用e记录下当前元素                if ((e = oldTab[j]) != null) {                    //将老数组j桶位置为空，方便回收                    oldTab[j] = null;                    //如果e节点不存在下一个节点，说明e是单个元素，则直接放置在新数组的桶位                    if (e.next == null)                        newTab[e.hash & (newCap - 1)] = e;                    //如果e是树节点，证明该节点处于红黑树中                    else if (e instanceof TreeNode)                        ((TreeNode)e).split(this, newTab, j, oldCap);                    //e为链表节点，则对链表进行遍历                    else { // preserve order                        //低位链表：存放在扩容之后的数组的下标位置，与当前数组下标位置一致                        //loHead：低位链表头节点                        //loTail低位链表尾节点                        Node loHead = null, loTail = null;                        //高位链表，存放扩容之后的数组的下标位置，=原索引+扩容之前数组容量                        //hiHead:高位链表头节点                        //hiTail:高位链表尾节点                        Node hiHead = null, hiTail = null;                        Node next;                        do {                            next = e.next;                            //oldCap为16:10000，与e.hsah做&运算可以得到高位为1还是0                            //高位为0，放在低位链表                            if ((e.hash & oldCap) == 0) {                                if (loTail == null)                                    //loHead指向e                                    loHead = e;                                else                                    loTail.next = e;                                loTail = e;                            }                            //高位为1，放在高位链表                            else {                                if (hiTail == null)                                    hiHead = e;                                else                                    hiTail.next = e;                                hiTail = e;                            }                        } while ((e = next) != null);                        //低位链表已成，将头节点loHead指向在原位                        if (loTail != null) {                            loTail.next = null;                            newTab[j] = loHead;                        }                        //高位链表已成，将头节点指向新索引                        if (hiTail != null) {                            hiTail.next = null;                            newTab[j + oldCap] = hiHead;                        }                    }                }            }        }        return newTab;    }

 * Because TreeNodes are about twice the size of regular nodes, we * use them only when bins contain enough nodes to warrant use * (see TREEIFY_THRESHOLD). And when they become too small (due to * removal or resizing) they are converted back to plain bins.  In * usages with well-distributed user hashCodes, tree bins are * rarely used.  Ideally, under random hashCodes, the frequency of * nodes in bins follows a Poisson distribution * (http://en.wikipedia.org/wiki/Poisson_distribution) with a * parameter of about 0.5 on average for the default resizing * 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)).

因为树节点的大小大约是普通节点的两倍，所以我们只在箱子包含足够的节点时才使用树节点（参见TREEIFY_THRESHOLD）。当他们边的太小（由于删除或调整大小）时，就会被转换回普通的桶，在使用分布良好的hashcode时，很少使用树箱。理想情况下，在随机哈希码下，箱子中节点的频率服从泊松分布第一个值是：
 * 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