hashMap常见面试题总览

• 为什么重写Equals还要重写HashCode方法？
• HashMap如何避免内存泄漏问题？
• HashMap1.7底层是如何实现的？
• HashMapKey为null存放在什么位置？
• HashMap如何解决Hash冲突问题？
• HashMap底层采用单链表还是双链表？
• HashMap根据key查询的时间复杂度？
• HashMap如何实现数组扩容问题？
• HashMap底层是有序存放的吗？
• LinkedHashMap 和 TreeMap底层如何实现有序的？
• HashMap底层如何降低Hash冲突概率？
• HashMap中hash函数是怎么实现的？
• 为什么不直接将key作为哈希值而是与高16位做异或运算？
• HashMap如何存放1万条key效率最高？
• HashMap高低位与取模运算有那些好处？
• HashMap1.8如何避免多线程扩容死循环问题？
• 为什么加载因子是0.75而不是1？
• 为什么HashMap1.8需要引入红黑树？
• 为什么链表长度>8需要转红黑树？而不是6？
• 什么情况下，需要从红黑树转换成链表存放？
• HashMap底层如何降低Hash冲突概率？
• 如何在高并发的情况下使用HashMap？
• ConcurrentHashMap底层实现的原理？

为什么重写equals还要重写hashCode方法？

HashMap如何避免内存泄漏问题？

• 回答
  • 为什么重写Equals还要重写HashCode方法？
  • HashMap如何避免内存泄漏问题？

Hashcode方法：底层采用C语言编写，根据对象内存地址转换成整数类型

 public native int hashCode();

基础1: 两个对象的hashCode值相等，对象不一定相等**

        String a1="a";Integer a2=97;//97 97System.out.println(a1.hashCode());System.out.println(a2.hashCode());

引发hash碰撞问题，所以还要重写Equals，String类型的Equals方法已经帮我们重写，所以下面比较是不相等的

        String a1="a";Integer a2=97;//falseSystem.out.println(a1.equals(a2));

String类型的Equals

    /*** Compares this string to the specified object.  The result is {@code* true} if and only if the argument is not {@code null} and is a {@code* String} object that represents the same sequence of characters as this* object.** @param  anObject*         The object to compare this {@code String} against** @return  {@code true} if the given object represents a {@code String}*          equivalent to this string, {@code false} otherwise** @see  #compareTo(String)* @see  #equalsIgnoreCase(String)*/public boolean equals(Object anObject) {if (this == anObject) {return true;}if (anObject instanceof String) {String anotherString = (String)anObject;int n = value.length;if (n == anotherString.value.length) {char v1[] = value;char v2[] = anotherString.value;int i = 0;while (n-- != 0) {if (v1[i] != v2[i])return false;i++;}return true;}}return false;}

Integer类型的Equals

 /*** Compares this object to the specified object.  The result is* {@code true} if and only if the argument is not* {@code null} and is an {@code Integer} object that* contains the same {@code int} value as this object.** @param   obj   the object to compare with.* @return  {@code true} if the objects are the same;*          {@code false} otherwise.*/public boolean equals(Object obj) {if (obj instanceof Integer) {return value == ((Integer)obj).intValue();}return false;}

基础2:两个对象equals相等，hashCode值不一定相等

       UserEntity userEntity = new UserEntity("wei", 18);UserEntity userEntity1 = new UserEntity("wei", 18);//false 内存地址肯定不相等System.out.println(userEntity==userEntity1);//false Object默认实现了equals方法，本质还是==，比较内存地址System.out.println(userEntity.equals(userEntity1));

Object类的equals

  public boolean equals(Object obj) {return (this == obj);}

结论：重写equals必须重写hashCode方法。

阿里规范
1.只要重写equals必须重写hashCode方法【强制】
2.Set、Map底层用hashCode和equals进行判断相等
3.String已重写equals和hashCode，所以可以用字符串作为Key【推荐】

不重写equals必须重写hashCode方法会导致内存无法回收（内存泄漏问题），同一个对象，一直创建新的空间存放。

HashMap1.7底层是如何实现的？

HashMapKey为null存放在什么位置？

基础：HashMap与HashTable之间的区别

HashMap底层如何实现

1.ArrayList集合实现，不需要考虑hash碰撞，时间复杂度为O（n）：
2. JDK1.7基于数组和链表(h=（key.hashCode)^h>>>16），不发生冲突为O（1），发生冲突链表为O（n），hashCode相同，对象不同
3. JDK1.8基于数组和链表和红黑树

源码解读

1.默认参数

 /*** The default initial capacity - MUST be a power of two.*///hashmap默认初始大小16,用位移做运算static final int DEFAULT_INITIAL_CAPACITY = 1 << 4; // aka 16/*** The maximum capacity, used if a higher value is implicitly specified* by either of the constructors with arguments.* MUST be a power of two <= 1<<30.*///2^30次方static final int MAXIMUM_CAPACITY = 1 << 30;/*** The load factor used when none specified in constructor.*///加载因子static final float DEFAULT_LOAD_FACTOR = 0.75f;/*** The bin count threshold for using a tree rather than list for a* bin.  Bins are converted to trees when adding an element to a* bin with at least this many nodes. The value must be greater* than 2 and should be at least 8 to mesh with assumptions in* tree removal about conversion back to plain bins upon* shrinkage.*///链表大于8，转换成红黑树static final int TREEIFY_THRESHOLD = 8;/*** The bin count threshold for untreeifying a (split) bin during a* resize operation. Should be less than TREEIFY_THRESHOLD, and at* most 6 to mesh with shrinkage detection under removal.*///链表小于6，转换成红黑树static final int UNTREEIFY_THRESHOLD = 6;/*** The smallest table capacity for which bins may be treeified.* (Otherwise the table is resized if too many nodes in a bin.)* Should be at least 4 * TREEIFY_THRESHOLD to avoid conflicts* between resizing and treeification thresholds.*///数组长度大于64，且链表长度大于8才会转换成红黑树static final int MIN_TREEIFY_CAPACITY = 64;

2.内部静态类

  static class Node<K,V> implements Map.Entry<K,V> {final int hash;final K key;V value;Node<K,V> next;Node(int hash, K key, V value, Node<K,V> next) {this.hash = hash;this.key = key;this.value = value;this.next = next;}public final K getKey()        { return key; }public final V getValue()      { return value; }public final String toString() { return key + "=" + value; }public final int hashCode() {return Objects.hashCode(key) ^ Objects.hashCode(value);}public final V setValue(V newValue) {V oldValue = value;value = newValue;return oldValue;}public final boolean equals(Object o) {if (o == this)return true;if (o instanceof Map.Entry) {Map.Entry<?,?> e = (Map.Entry<?,?>)o;if (Objects.equals(key, e.getKey()) &&Objects.equals(value, e.getValue()))return true;}return false;}}

3.字段

 /* ---------------- Fields -------------- *//*** The table, initialized on first use, and resized as* necessary. When allocated, length is always a power of two.* (We also tolerate length zero in some operations to allow* bootstrapping mechanics that are currently not needed.)*///数组 transient不能序列化transient Node<K,V>[] table;/*** Holds cached entrySet(). Note that AbstractMap fields are used* for keySet() and values().*/transient Set<Map.Entry<K,V>> entrySet;/*** The number of key-value mappings contained in this map.*/transient int size;/*** The number of times this HashMap has been structurally modified* Structural modifications are those that change the number of mappings in* the HashMap or otherwise modify its internal structure (e.g.,* rehash).  This field is used to make iterators on Collection-views of* the HashMap fail-fast.  (See ConcurrentModificationException).*/// fail-fast机制transient int modCount;/*** The next size value at which to resize (capacity * load factor).** @serial*/// (The javadoc description is true upon serialization.// Additionally, if the table array has not been allocated, this// field holds the initial array capacity, or zero signifying// DEFAULT_INITIAL_CAPACITY.)int threshold;/*** The load factor for the hash table.** @serial*///加载因子，如果不填写则采用默认加载因子final float loadFactor;

4.forEach修改报错

       public final void forEach(Consumer<? super K> action) {Node<K,V>[] tab;if (action == null)throw new NullPointerException();if (size > 0 && (tab = table) != null) {int mc = modCount;for (int i = 0; i < tab.length; ++i) {for (Node<K,V> e = tab[i]; e != null; e = e.next)action.accept(e.key);}if (modCount != mc)throw new ConcurrentModificationException();}}}

5.put方法

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

    static final int hash(Object key) {int h;// null值放在0号位置，科学方法计算hash值，减少冲突return (key == null) ? 0 : (h = key.hashCode()) ^ (h >>> 16);}

  /*** Implements Map.put and related methods** @param hash hash for key* @param key the key* @param value the value to put* @param onlyIfAbsent if true, don't change existing value 当不存在时，则不更改现有值* @param evict if false, the table is in creation mode. 如果为false，则退出，表处于创建模式* @return previous value, or null if none*/final V putVal(int hash, K key, V value, boolean onlyIfAbsent,boolean evict) {//数组+链表 n 数组长度 i链表存放位置Node<K,V>[] tab; Node<K,V> p; int n, i;//数组不存在或者数组长度为0，则调用扩容方法，懒加载if ((tab = table) == null || (n = tab.length) == 0)n = (tab = resize()).length;//hash不冲突，数组直接存放链表if ((p = tab[i = (n - 1) & hash]) == null)tab[i] = newNode(hash, key, value, null);else {Node<K,V> e; K k;//如果hash值相等且key值相等，将p赋值给e，等待做更新操作if (p.hash == hash &&((k = p.key) == key || (key != null && key.equals(k))))e = p;//是红黑树else if (p instanceof TreeNode)e = ((TreeNode<K,V>)p).putTreeVal(this, tab, hash, key, value);else {for (int binCount = 0; ; ++binCount) {// 遍历链表，到末尾添加元素if ((e = p.next) == null) {p.next = newNode(hash, key, value, null);//帮链表转换成红黑树，数组容量>=64 && 链表长度大于8if (binCount >= TREEIFY_THRESHOLD - 1) // -1 for 1sttreeifyBin(tab, hash);break;}if (e.hash == hash &&((k = e.key) == key || (key != null && key.equals(k))))break;p = e;}}// 存在则直接修改if (e != null) { // existing mapping for keyV oldValue = e.value;if (!onlyIfAbsent || oldValue == null)e.value = value;afterNodeAccess(e);return oldValue;}}++modCount;if (++size > threshold)resize();afterNodeInsertion(evict);return null;}

扩容方法

  /*** 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*/final Node<K,V>[] resize() {//旧的数组Node<K,V>[] oldTab = table;//获取旧数组的长度int oldCap = (oldTab == null) ? 0 : oldTab.length;// 旧数组的扩容阈值int oldThr = threshold;// 新的数组长度和新的扩容阈值int newCap, newThr = 0;// 如果旧的数组存在if (oldCap > 0) {if (oldCap >= MAXIMUM_CAPACITY) {threshold = Integer.MAX_VALUE;return oldTab;}else if ((newCap = oldCap << 1) < MAXIMUM_CAPACITY &&oldCap >= DEFAULT_INITIAL_CAPACITY)newThr = oldThr << 1; // double threshold 两倍扩容}// 旧数组存在，将旧的数组的长度赋值给新数组else if (oldThr > 0) // initial capacity was placed in thresholdnewCap = oldThr;//使用默认值进行初始化else {               // zero initial threshold signifies using defaults//初始化数组容量=数组默认初始大小16newCap = DEFAULT_INITIAL_CAPACITY;// 初始化数组扩容阈值=链表加载因子*数组默认初始大小12newThr = (int)(DEFAULT_LOAD_FACTOR * DEFAULT_INITIAL_CAPACITY);}//如果新数组扩容阈值为0if (newThr == 0) {//ft = 新的数组长度*加载因子，即计算阈值float ft = (float)newCap * loadFactor; newThr = (newCap < MAXIMUM_CAPACITY && ft < (float)MAXIMUM_CAPACITY ?(int)ft : Integer.MAX_VALUE);}//将新数组扩容阈值赋值到全局变量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//扩容是2的次方，分高位低位进行移动操作，插入新数组，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;elseloTail.next = e;loTail = e;}else {if (hiTail == null)hiHead = e;elsehiTail.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;}