HashMap1.8常见面试问题
1.hashmap转红黑树的时机：
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); break; }
上面代码在putval方法中：判断了当不存在的key插入map时，如果往node节点一直遍历当大于等于8次时，会转化尝试转化为红黑树，注意是尝试。
在treeifyBin方法中还会判断：
在treeifyBin方法中还会判断：
if (tab == null || (n = tab.length) < MIN_TREEIFY_CAPACITY) resize() ;
当tab的长度也就是cap容量小于64时会resize而不是直接转化为红黑树.
综上：hashmap转红黑树的时机为当链表长度大于等于8并且容量大于等于64时。
2.hashmap put过程
源码如下：
final V putVal(int hash, K key, V value, boolean onlyIfAbsent, boolean evict) { //tab为node数组，p为hash运算后的下标i的node节点，n为tab长度，i为node数组下标 Node<K,V>[] tab; Node<K,V> p; int n, i; //hashmap采用懒加载思想，只有真正用的时候才会初始化 if ((tab = table) == null || (n = tab.length) == 0) n = (tab = resize()).length; //里面有个知识点 运算符的优先级 ()>&>= 所以先用长度-1和hash值与运算再赋给坐标i if ((p = tab[i = (n - 1) & hash]) == null) //tab[i]空直接插入 tab[i] = newNode(hash, key, value, null); else { //e为后面要操作的node节点 k为node的key p为已存在的下标i的node节点 Node<K,V> e; K k; //常问的重写hashcode和equals方法答案就在下面这句话 //当p和e的hash值相同并且引用p,k相同或者key调用equals方法相同时直接覆盖 if (p.hash == hash && ((k = p.key) == key || (key != null && key.equals(k)))) //将把p赋给e，后续e都是我们默认操作的put的key的节点 e = p; else if (p instanceof TreeNode) //如果是树则使用其他的put方法 e = ((TreeNode<K,V>)p).putTreeVal(this, tab, hash, key, value); else { //下面是链表的遍历过程 for (int binCount = 0; ; ++binCount) { if ((e = p.next) == null) { p.next = newNode(hash, key, value, null); //尝试转红黑树 if (binCount >= TREEIFY_THRESHOLD - 1) // -1 for 1st treeifyBin(tab, hash); break; } //找到存在的key 跳出循环 if (e.hash == hash && ((k = e.key) == key || (key != null && key.equals(k)))) break; p = e; } } //上面链表遍历有3种结果 1.node链表长度小于8没有找到对应key，直接尾插法new一个node 2.node链表大于8，尝试红黑树转换 3.node链表长度小于8找到对应的key，跳出循环，此时e是有值 所以我们说e是最后我们操作存在key的node节点 //返回旧值，放入新值 if (e != null) { // existing mapping for key V oldValue = e.value; if (!onlyIfAbsent || oldValue == null) e.value = value; afterNodeAccess(e); return oldValue; } } ++modCount; //是否扩容 if (++size > threshold) resize(); afterNodeInsertion(evict); return null; }
3.hashmap resize过程
resize总结：1.获取新的容量和临界值
2.创建新表将hashmap tab rehash 元素以原来下标或者二次幂的偏移量下标移动
/** * 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 */
//二次幂展开，扩容后元素要么是原有下标，要么以二次幂偏移量
//exp:假设容量为16 key的hash为16 1111&10000=0存放在tab[0]中 如果扩容后容量为32 那么rehash时11111&10000 就是tab[16] 所以以二次幂的偏移量移动 final Node<K,V>[] resize() { Node<K,V>[] oldTab = table;
//oldTab==null 为了兼容初始化 int oldCap = (oldTab == null) ? 0 : oldTab.length; int oldThr = threshold; int newCap, newThr = 0; if (oldCap > 0) {
//当容量大于允许最大容量时临界点为Integer最大值 if (oldCap >= MAXIMUM_CAPACITY) { threshold = Integer.MAX_VALUE; return oldTab; }
//正常扩容 临界点和容量都扩大两倍 else if ((newCap = oldCap << 1) < MAXIMUM_CAPACITY && oldCap >= DEFAULT_INITIAL_CAPACITY) newThr = oldThr << 1; // double threshold }//使用临界值覆盖容量 else if (oldThr > 0) // initial capacity was placed in threshold newCap = oldThr; else { // zero initial threshold signifies using defaults //当oldThr是0时初始化
newCap = DEFAULT_INITIAL_CAPACITY; newThr = (int)(DEFAULT_LOAD_FACTOR * DEFAULT_INITIAL_CAPACITY); }
//临界值还是为0 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"})
//创建新tab Node<K,V>[] newTab = (Node<K,V>[])new Node[newCap]; table = newTab; if (oldTab != null) {
//遍历oldTab数组 for (int j = 0; j < oldCap; ++j) { Node<K,V> e; if ((e = oldTab[j]) != null) { oldTab[j] = null; if (e.next == null)
//该下标只有一个node时直接放入新tab中 newTab[e.hash & (newCap - 1)] = e; else if (e instanceof TreeNode) ((TreeNode<K,V>)e).split(this, newTab, j, oldCap); else { // preserve order
//下面的代码分为了两个链表 其实就是把hash大于等于cap的和小于的分开 小于的在原位 大于等于的按照oldcap偏移量下标存放
//并且采用了尾插法
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; }
