Java面试一天一题（day 3面试题：HashMap1.8相比1.7做了哪些变更，它是线程安全的吗?）

static final class TreeNode<K,V> extends LinkedHashMap.Entry<K,V> {  
      //父节点、左子树、右子树、删除辅助节点 + 颜色
    TreeNode<K,V> parent;  
    TreeNode<K,V> left;   
    TreeNode<K,V> right;
    TreeNode<K,V> prev;   
    boolean red;   
    TreeNode(int hash, K key, V val, Node<K,V> next) {  
        super(hash, key, val, next);  
    }  
    // 返回当前节点的根节点  
    final TreeNode<K,V> root() {  
        for (TreeNode<K,V> r = this, p;;) {  
            if ((p = r.parent) == null)  
                return r;  
            r = p;  
        }  
}

//1.7为Entry对象存储k，v的数组，1.8改为Node类型数组，长度均为2的幂次方数
transient Node<K,V>[] table;
//桶的树化阈值：即 链表转成红黑树的阈值，在存储数据时，当链表长度 > 8时，则将链表转换成红黑树
static final int TREEIFY_THRESHOLD = 8; 
//桶的链表还原阈值：即 红黑树转为链表的阈值，当在扩容（resize（））时（此时HashMap的数据存储位置会重新计算），在重新计算存储位置后，当原有的红黑树内数量 < 6时，则将红黑树转换成链表
static final int UNTREEIFY_THRESHOLD = 6;
//最小树形化容量阈值：即 当哈希表中的容量 > 该值时，才允许树形化链表 （即 将链表 转换成红黑树）
// 否则，若桶内元素太多时，则直接扩容，而不是树形化
// 为了避免进行扩容、树形化选择的冲突，这个值不能小于 4 * TREEIFY_THRESHOLD
static final int MIN_TREEIFY_CAPACITY = 64;

public V get(Object key) {
        Node<K,V> e;
        // 1. 计算key的hash值
        // 2. 通过getNode（）获取所查询的数据
        // 3. 获取后，判断数据是否为空
        return (e = getNode(hash(key), key)) == null ? null : e.value;
    }
     
final Node<K,V> getNode(int hash, Object key) {
    Node<K,V>[] tab; Node<K,V> first, e; int n; K k;


    // 1. 计算存放在数组table中的位置
    if ((tab = table) != null && (n = tab.length) > 0 &&
        (first = tab[(n - 1) & hash]) != null) {


        // 2. 在数组、红黑树、链表中依次查找，用equals判断
        // 2.1 先在数组中找，若存在，则直接返回
        if (first.hash == hash && // always check first node
            ((k = first.key) == key || (key != null && key.equals(k))))
            return first;


        // 2.2. 若数组中没有，则到红黑树中寻找
        if ((e = first.next) != null) {
            // 在树中get
            if (first instanceof TreeNode)
                return ((TreeNode<K,V>)first).getTreeNode(hash, key);


            // 2.3. 若红黑树中也没有，则遍历链表
            do {
                if (e.hash == hash &&
                    ((k = e.key) == key || (key != null && key.equals(k))))
                    return e;
            } while ((e = e.next) != null);
        }
    }
    return null;
}

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;
  //1.第一次put，若table为空，此时会通过resize()初始化
  if ((tab = table) == null || (n = tab.length) == 0)
      n = (tab = resize()).length;
  //2.计算下标
  if ((p = tab[i = (n - 1) & hash]) == null)
      tab[i] = newNode(hash, key, value, null);
  else {
      Node<K,V> e; K k;
      if (p.hash == hash &&
          ((k = p.key) == key || (key != null && key.equals(k))))
          e = p;
      //3.是红黑树or链表
      else if (p instanceof TreeNode)
          e = ((TreeNode<K,V>)p).putTreeVal(this, tab, hash, key, value);
      else {
      //4.遍历链表，通过equals判断是直接插入还是覆盖更新
      //新增节点后需要判断需要判断链表长度是否超过8（8为链表树化阈值），超过8则转化为红黑树
          for (int binCount = 0; ; ++binCount) {
              if ((e = p.next) == null) {
                  p.next = newNode(hash, key, value, null);
                  if (binCount >= TREEIFY_THRESHOLD - 1) // -1 for 1st
                      treeifyBin(tab, hash); //转化为红黑树,从0开始遍历所以是>=7
                  break;
              }
              if (e.hash == hash &&
                  ((k = e.key) == key || (key != null && key.equals(k))))
                  break;
              p = e;
          }
      }
      if (e != null) { // existing mapping for key
          V oldValue = e.value;
          if (!onlyIfAbsent || oldValue == null)
              e.value = value; //替换老值
          afterNodeAccess(e);
          return oldValue;
      }
  }
  ++modCount;
  if (++size > threshold)
      resize();//5.扩容
  afterNodeInsertion(evict);
  return null;
}

 final Node<K,V>[] resize() {
   Node<K,V>[] oldTab = table;  //扩容前数组
   int oldCap = (oldTab == null) ? 0 : oldTab.length; //扩容前数组容量
   int oldThr = threshold;  //扩容前数组阈值
   int newCap, newThr = 0;  
   if (oldCap > 0) {
   //超过最大容量，不扩容
       if (oldCap >= MAXIMUM_CAPACITY) {
           threshold = Integer.MAX_VALUE;
           return oldTab;
       }//容量不够，需要扩容
       else if ((newCap = oldCap << 1) < MAXIMUM_CAPACITY &&
                oldCap >= DEFAULT_INITIAL_CAPACITY)
           newThr = oldThr << 1; // double threshold   左移扩容两倍
   }
   else if (oldThr > 0) // initial capacity was placed in threshold
       newCap = oldThr;
   else {               // zero initial threshold signifies using defaults初始化新table
       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;
   //移动old bucket到新的桶
   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;
                       //原索引
                       if ((e.hash & oldCap) == 0) {
                           if (loTail == null)
                               loHead = e;
                           else
                               loTail.next = e;
                           loTail = e;
                       }//原索引 + oldCap
                       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;
}

解释：比如key.hashcode = 10010110101010,那么如何直接拿 h & （len-1） 此处len-1 = 15
   10010110101010
            01111
-----------------------
            01010
此时是否发现h的高16位压根就没有进行计算，那么只用低位计算是不是冲突的概率较大？
这就是为什么将h=key.hashcode的高位和低位进行^操作后在与len-1进行&操作