面试28k职位，老乡面试官从HashCode到HashMap给我讲了一下午！

// 获取hashCode "abc".hashCode();
public int hashCode() {
    int h = hash;
    if (h == 0 && value.length > 0) {
        char val[] = value;
        for (int i = 0; i < value.length; i++) {
            h = 31 * h + val[i];
        }
        hash = h;
    }
    return h;
}

The value 31 was chosen because it is an odd prime. If it were even and the multiplication overflowed, information would be lost, as multiplication by 2 is equivalent to shifting. The advantage of using a prime is less clear, but it is traditional. A nice property of 31 is that the multiplication can be replaced by a shift and a subtraction for better performance: 31 * i == (i << 5) - i. Modern VMs do this sort of optimization automatically.

As Goodrich and Tamassia point out, If you take over 50,000 English words (formed as the union of the word lists provided in two variants of Unix), using the constants 31, 33, 37, 39, and 41 will produce less than 7 collisions in each case. Knowing this, it should come as no surprise that many Java implementations choose one of these constants.

1	a	"n.(A)As 或 A's  安(ampere(a) art.一;n.字母A /[军] Analog.Digital,模拟/数字 /(=account of) 帐上"
2	aaal	American Academy of Arts and Letters 美国艺术和文学学会
3	aachen	 亚琛[德意志联邦共和国西部城市]
4	aacs	Airways and Air Communications Service (美国)航路与航空通讯联络处
5	aah	" [军]Armored Artillery Howitzer,装甲榴弹炮;[军]Advanced Attack Helicopter,先进攻击直升机"
6	aal	"ATM Adaptation Layer,ATM适应层"
7	aapamoor	"n.[生]丘泽,高低位镶嵌沼泽"

public static Integer hashCode(String str, Integer multiplier) {
    int hash = 0;
    for (int i = 0; i < str.length(); i++) {
        hash = multiplier * hash + str.charAt(i);
    }
    return hash;
}

private static RateInfo hashCollisionRate(Integer multiplier, List<Integer> hashCodeList) {
    int maxHash = hashCodeList.stream().max(Integer::compareTo).get();
    int minHash = hashCodeList.stream().min(Integer::compareTo).get();
    int collisionCount = (int) (hashCodeList.size() - hashCodeList.stream().distinct().count());
    double collisionRate = (collisionCount * 1.0) / hashCodeList.size();
    return new RateInfo(maxHash, minHash, multiplier, collisionCount, collisionRate);
}

@Before
public void before() {
    "abc".hashCode();
    // 读取文件，103976个英语单词库.txt
    words = FileUtil.readWordList("E:/itstack/git/github.com/interview/interview-01/103976个英语单词库.txt");
}

@Test
public void test_collisionRate() {
    List<RateInfo> rateInfoList = HashCode.collisionRateList(words, 2, 3, 5, 7, 17, 31, 32, 33, 39, 41, 199);
    for (RateInfo rate : rateInfoList) {
        System.out.println(String.format("乘数 = %4d, 最小Hash = %11d, 最大Hash = %10d, 碰撞数量 =%6d, 碰撞概率 = %.4f%%", rate.getMultiplier(), rate.getMinHash(), rate.getMaxHash(), rate.getCollisionCount(), rate.getCollisionRate() * 100));
    }
}

单词数量：103976
乘数 =    2, 最小Hash =          97, 最大Hash = 1842581979, 碰撞数量 = 60382, 碰撞概率 = 58.0730%
乘数 =    3, 最小Hash = -2147308825, 最大Hash = 2146995420, 碰撞数量 = 24300, 碰撞概率 = 23.3708%
乘数 =    5, 最小Hash = -2147091606, 最大Hash = 2147227581, 碰撞数量 =  7994, 碰撞概率 = 7.6883%
乘数 =    7, 最小Hash = -2147431389, 最大Hash = 2147226363, 碰撞数量 =  3826, 碰撞概率 = 3.6797%
乘数 =   17, 最小Hash = -2147238638, 最大Hash = 2147101452, 碰撞数量 =   576, 碰撞概率 = 0.5540%
乘数 =   31, 最小Hash = -2147461248, 最大Hash = 2147444544, 碰撞数量 =     2, 碰撞概率 = 0.0019%
乘数 =   32, 最小Hash = -2007883634, 最大Hash = 2074238226, 碰撞数量 = 34947, 碰撞概率 = 33.6106%
乘数 =   33, 最小Hash = -2147469046, 最大Hash = 2147378587, 碰撞数量 =     1, 碰撞概率 = 0.0010%
乘数 =   39, 最小Hash = -2147463635, 最大Hash = 2147443239, 碰撞数量 =     0, 碰撞概率 = 0.0000%
乘数 =   41, 最小Hash = -2147423916, 最大Hash = 2147441721, 碰撞数量 =     1, 碰撞概率 = 0.0010%
乘数 =  199, 最小Hash = -2147459902, 最大Hash = 2147480320, 碰撞数量 =     0, 碰撞概率 = 0.0000%

Process finished with exit code 0

public static Map<Integer, Integer> hashArea(List<Integer> hashCodeList) {
    Map<Integer, Integer> statistics = new LinkedHashMap<>();
    int start = 0;
    for (long i = 0x80000000; i <= 0x7fffffff; i += 67108864) {
        long min = i;
        long max = min + 67108864;
        // 筛选出每个格子里的哈希值数量，java8流统计；https://bugstack.cn/itstack-demo-any/2019/12/10/%E6%9C%89%E7%82%B9%E5%B9%B2%E8%B4%A7-Jdk1.8%E6%96%B0%E7%89%B9%E6%80%A7%E5%AE%9E%E6%88%98%E7%AF%87(41%E4%B8%AA%E6%A1%88%E4%BE%8B).html
        int num = (int) hashCodeList.parallelStream().filter(x -> x >= min && x < max).count();
        statistics.put(start++, num);
    }
    return statistics;

@Test
public void test_hashArea() {
    System.out.println(HashCode.hashArea(words, 2).values());
    System.out.println(HashCode.hashArea(words, 7).values());
    System.out.println(HashCode.hashArea(words, 31).values());
    System.out.println(HashCode.hashArea(words, 32).values());
    System.out.println(HashCode.hashArea(words, 199).values());
}

// 初始化一组字符串
List<String> list = new ArrayList<>();
list.add("jlkk");
list.add("lopi");
list.add("小傅哥");
list.add("e4we");
list.add("alpo");
list.add("yhjk");
list.add("plop");

// 定义要存放的数组
String[] tab = new String[8];

// 循环存放
for (String key : list) {
    int idx = key.hashCode() & (tab.length - 1);  // 计算索引位置
    System.out.println(String.format("key值=%s Idx=%d", key, idx));
    if (null == tab[idx]) {
        tab[idx] = key;
        continue;
    }
    tab[idx] = tab[idx] + "->" + key;
}
// 输出测试结果
System.out.println(JSON.toJSONString(tab));

key值=jlkk Idx=2
key值=lopi Idx=4
key值=小傅哥 Idx=7
key值=e4we Idx=5
key值=alpo Idx=2
key值=yhjk Idx=0
key值=plop Idx=5
测试结果：["yhjk",null,"jlkk->alpo",null,"lopi","e4we->plop",null,"小傅哥"]

static final int hash(Object key) {
    int h;
    return (key == null) ? 0 : (h = key.hashCode()) ^ (h >>> 16);
}

public class Disturb {

    public static int disturbHashIdx(String key, int size) {
        return (size - 1) & (key.hashCode() ^ (key.hashCode() >>> 16));
    }

    public static int hashIdx(String key, int size) {
        return (size - 1) & key.hashCode();
    }

}

// 10万单词已经初始化到words中
@Test
public void test_disturb() {
    Map<Integer, Integer> map = new HashMap<>(16);
    for (String word : words) {
        // 使用扰动函数
        int idx = Disturb.disturbHashIdx(word, 128);
        // 不使用扰动函数
        // int idx = Disturb.hashIdx(word, 128);
        if (map.containsKey(idx)) {
            Integer integer = map.get(idx);
            map.put(idx, ++integer);
        } else {
            map.put(idx, 1);
        }
    }
    System.out.println(map.values());
}

public HashMap(int initialCapacity, float loadFactor) {
    ...
    this.loadFactor = loadFactor;
    this.threshold = tableSizeFor(initialCapacity);
}

static final int tableSizeFor(int cap) {
    int n = cap - 1;
    n |= n >>> 1;
    n |= n >>> 2;
    n |= n >>> 4;
    n |= n >>> 8;
    n |= n >>> 16;
    return (n < 0) ? 1 : (n >= MAXIMUM_CAPACITY) ? MAXIMUM_CAPACITY : n + 1;
}

static final float DEFAULT_LOAD_FACTOR = 0.75f;

@Test
public void test_hashMap() {
    List<String> list = new ArrayList<>();
    list.add("jlkk");
    list.add("lopi");
    list.add("jmdw");
    list.add("e4we");
    list.add("io98");
    list.add("nmhg");
    list.add("vfg6");
    list.add("gfrt");
    list.add("alpo");
    list.add("vfbh");
    list.add("bnhj");
    list.add("zuio");
    list.add("iu8e");
    list.add("yhjk");
    list.add("plop");
    list.add("dd0p");
    for (String key : list) {
        int hash = key.hashCode() ^ (key.hashCode() >>> 16);
        System.out.println("字符串：" + key + " \tIdx(16)：" + ((16 - 1) & hash) + " \tBit值：" + Integer.toBinaryString(hash) + " - " + Integer.toBinaryString(hash & 16) + " \t\tIdx(32)：" + ((
        System.out.println(Integer.toBinaryString(key.hashCode()) +" "+ Integer.toBinaryString(hash) + " " + Integer.toBinaryString((32 - 1) & hash));
    }
}

字符串：jlkk 	Idx(16)：3 	Bit值：1100011101001000010011 - 10000 		Idx(32)：19
1100011101001000100010 1100011101001000010011 10011
字符串：lopi 	Idx(16)：14 	Bit值：1100101100011010001110 - 0 		Idx(32)：14
1100101100011010111100 1100101100011010001110 1110
字符串：jmdw 	Idx(16)：7 	Bit值：1100011101010100100111 - 0 		Idx(32)：7
1100011101010100010110 1100011101010100100111 111
字符串：e4we 	Idx(16)：3 	Bit值：1011101011101101010011 - 10000 		Idx(32)：19
1011101011101101111101 1011101011101101010011 10011
字符串：io98 	Idx(16)：4 	Bit值：1100010110001011110100 - 10000 		Idx(32)：20
1100010110001011000101 1100010110001011110100 10100
字符串：nmhg 	Idx(16)：13 	Bit值：1100111010011011001101 - 0 		Idx(32)：13
1100111010011011111110 1100111010011011001101 1101
字符串：vfg6 	Idx(16)：8 	Bit值：1101110010111101101000 - 0 		Idx(32)：8
1101110010111101011111 1101110010111101101000 1000
字符串：gfrt 	Idx(16)：1 	Bit值：1100000101111101010001 - 10000 		Idx(32)：17
1100000101111101100001 1100000101111101010001 10001
字符串：alpo 	Idx(16)：7 	Bit值：1011011011101101000111 - 0 		Idx(32)：7
1011011011101101101010 1011011011101101000111 111
字符串：vfbh 	Idx(16)：1 	Bit值：1101110010111011000001 - 0 		Idx(32)：1
1101110010111011110110 1101110010111011000001 1
字符串：bnhj 	Idx(16)：0 	Bit值：1011100011011001100000 - 0 		Idx(32)：0
1011100011011001001110 1011100011011001100000 0
字符串：zuio 	Idx(16)：8 	Bit值：1110010011100110011000 - 10000 		Idx(32)：24
1110010011100110100001 1110010011100110011000 11000
字符串：iu8e 	Idx(16)：8 	Bit值：1100010111100101101000 - 0 		Idx(32)：8
1100010111100101011001 1100010111100101101000 1000
字符串：yhjk 	Idx(16)：8 	Bit值：1110001001010010101000 - 0 		Idx(32)：8
1110001001010010010000 1110001001010010101000 1000
字符串：plop 	Idx(16)：9 	Bit值：1101001000110011101001 - 0 		Idx(32)：9
1101001000110011011101 1101001000110011101001 1001
字符串：dd0p 	Idx(16)：14 	Bit值：1011101111001011101110 - 0 		Idx(32)：14
1011101111001011000000 1011101111001011101110 1110

   n = (tab = resize()).length;

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;
    // 初始化桶数组 table，table 被延迟到插入新数据时再进行初始化
    if ((tab = table) == null || (n = tab.length) == 0)
        n = (tab = resize()).length;
    // 如果桶中不包含键值对节点引用，则将新键值对节点的引用存入桶中即可
    if ((p = tab[i = (n - 1) & hash]) == null)
        tab[i] = newNode(hash, key, value, null);
    else {
        Node<K,V> e; K k;
        // 如果键的值以及节点 hash 等于链表中的第一个键值对节点时，则将 e 指向该键值对
        if (p.hash == hash &&
            ((k = p.key) == key || (key != null && key.equals(k))))
            e = p;
            
        // 如果桶中的引用类型为 TreeNode，则调用红黑树的插入方法
        else if (p instanceof TreeNode)  
            e = ((TreeNode<K,V>)p).putTreeVal(this, tab, hash, key, value);
        else {
            // 对链表进行遍历，并统计链表长度
            for (int binCount = 0; ; ++binCount) {
                // 链表中不包含要插入的键值对节点时，则将该节点接在链表的最后
                if ((e = p.next) == null) {
                    p.next = newNode(hash, key, value, null);
                    // 如果链表长度大于或等于树化阈值，则进行树化操作
                    if (binCount >= TREEIFY_THRESHOLD - 1) // -1 for 1st
                        treeifyBin(tab, hash);
                    break;
                }
                
                // 条件为 true，表示当前链表包含要插入的键值对，终止遍历
                if (e.hash == hash &&
                    ((k = e.key) == key || (key != null && key.equals(k))))
                    break;
                p = e;
            }
        }
        
        // 判断要插入的键值对是否存在 HashMap 中
        if (e != null) { // existing mapping for key
            V oldValue = e.value;
            // onlyIfAbsent 表示是否仅在 oldValue 为 null 的情况下更新键值对的值
            if (!onlyIfAbsent || oldValue == null)
                e.value = value;
            afterNodeAccess(e);
            return oldValue;
        }
    }
    ++modCount;
    // 键值对数量超过阈值时，则进行扩容
    if (++size > threshold)
        resize();
    afterNodeInsertion(evict);
    return null;
}

final Node<K,V>[] resize() {
    Node<K,V>[] oldTab = table;
    int oldCap = (oldTab == null) ? 0 : oldTab.length;
    int oldThr = threshold;
    int newCap, newThr = 0;
    // Cap 是 capacity 的缩写，容量。如果容量不为空，则说明已经初始化。
    if (oldCap > 0) {
        // 如果容量达到最大1 << 30则不再扩容
        if (oldCap >= MAXIMUM_CAPACITY) {
            threshold = Integer.MAX_VALUE;
            return oldTab;
        }
        
        // 按旧容量和阀值的2倍计算新容量和阀值
        else if ((newCap = oldCap << 1) < MAXIMUM_CAPACITY &&
                 oldCap >= DEFAULT_INITIAL_CAPACITY)
            newThr = oldThr << 1; // double threshold
    }
    else if (oldThr > 0) // initial capacity was placed in threshold
    
        // initial capacity was placed in threshold 翻译过来的意思，如下；
        // 初始化时，将 threshold 的值赋值给 newCap，
        // HashMap 使用 threshold 变量暂时保存 initialCapacity 参数的值
        newCap = oldThr;
    else {               // zero initial threshold signifies using defaults
        // 这一部分也是，源代码中也有相应的英文注释
        // 调用无参构造方法时，数组桶数组容量为默认容量 1 << 4; aka 16
        // 阀值；是默认容量与负载因子的乘积，0.75
        newCap = DEFAULT_INITIAL_CAPACITY;
        newThr = (int)(DEFAULT_LOAD_FACTOR * DEFAULT_INITIAL_CAPACITY);
    }
    
    // newThr为0，则使用阀值公式计算容量
    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"})
        // 初始化数组桶，用于存放key
        Node<K,V>[] newTab = (Node<K,V>[])new Node[newCap];
    table = newTab;
    if (oldTab != null) {
        // 如果旧数组桶，oldCap有值，则遍历将键值映射到新数组桶中
        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)
                    // 这里split，是红黑树拆分操作。在重新映射时操作的。
                    ((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;
                    // 这里是链表，如果当前是按照链表存放的，则将链表节点按原顺序进行分组{这里有专门的文章介绍，如何不需要重新计算哈希值进行拆分《HashMap核心知识，扰动函数、负载因子、扩容链表拆分，深度学习》}
                    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);
                    
                    // 将分组后的链表映射到桶中
                    if (loTail != null) {
                        loTail.next = null;
                        newTab[j] = loHead;
                    }
                    if (hiTail != null) {
                        hiTail.next = null;
                        newTab[j + oldCap] = hiHead;
                    }
                }
            }
        }
    }
    return newTab;
}

final void treeifyBin(Node<K,V>[] tab, int hash) {
    int n, index; Node<K,V> e;
    // 这块就是我们上面提到的，不一定树化还可能只是扩容。主要桶数组容量是否小于64 MIN_TREEIFY_CAPACITY 
    if (tab == null || (n = tab.length) < MIN_TREEIFY_CAPACITY)
        resize();
    else if ((e = tab[index = (n - 1) & hash]) != null) {
    	// 又是单词缩写；hd = head (头部)，tl = tile (结尾)
        TreeNode<K,V> hd = null, tl = null;
        do {
            // 将普通节点转换为树节点，但此时还不是红黑树，也就是说还不一定平衡
            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)
            // 转红黑树操作，这里需要循环比较，染色、旋转。关于红黑树，在下一章节详细讲解
            hd.treeify(tab);
    }
}

final Node<K,V> untreeify(HashMap<K,V> map) {
    Node<K,V> hd = null, tl = null;
    // 遍历TreeNode
    for (Node<K,V> q = this; q != null; q = q.next) {
    	// TreeNode替换Node
        Node<K,V> p = map.replacementNode(q, null);
        if (tl == null)
            hd = p;
        else
            tl.next = p;
        tl = p;
    }
    return hd;
}

// 替换方法
Node<K,V> replacementNode(Node<K,V> p, Node<K,V> next) {
    return new Node<>(p.hash, p.key, p.value, next);
}

public V get(Object key) {
    Node<K,V> e;
    // 同样需要经过扰动函数计算哈希值
    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;
    // 判断桶数组的是否为空和长度值
    if ((tab = table) != null && (n = tab.length) > 0 &&
        // 计算下标，哈希值与数组长度-1
        (first = tab[(n - 1) & hash]) != null) {
        if (first.hash == hash && // always check first node
            ((k = first.key) == key || (key != null && key.equals(k))))
            return first;
        if ((e = first.next) != null) {
            // TreeNode 节点直接调用红黑树的查找方法，时间复杂度O(logn)
            if (first instanceof TreeNode)
                return ((TreeNode<K,V>)first).getTreeNode(hash, key);
            // 如果是链表就依次遍历查找
            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 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;
    // 定位桶数组中的下标位置，index = (n - 1) & hash
    if ((tab = table) != null && (n = tab.length) > 0 &&
        (p = tab[index = (n - 1) & hash]) != null) {
        Node<K,V> node = null, e; K k; V v;
        // 如果键的值与链表第一个节点相等，则将 node 指向该节点
        if (p.hash == hash &&
            ((k = p.key) == key || (key != null && key.equals(k))))
            node = p;
        else if ((e = p.next) != null) {
            // 树节点，调用红黑树的查找方法，定位节点。
            if (p instanceof TreeNode)
                node = ((TreeNode<K,V>)p).getTreeNode(hash, key);
            else {
                // 遍历链表，找到待删除节点
                do {
                    if (e.hash == hash &&
                        ((k = e.key) == key ||
                         (key != null && key.equals(k)))) {
                        node = e;
                        break;
                    }
                    p = e;
                } while ((e = e.next) != null);
            }
        }
        
        // 删除节点，以及红黑树需要修复，因为删除后会破坏平衡性。链表的删除更加简单。
        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;
            ++modCount;
            --size;
            afterNodeRemoval(node);
            return node;
        }
    }
    return null;
} 

 for (String key : map.keySet()) {
     System.out.print(key + " ");
 }

 for (HashMap.Entry entry : map.entrySet()) {
     System.out.print(entry + " ");
 }

@Test
public void test_Iterator() {
    Map<String, String> map = new HashMap<String, String>(64);
    map.put("24", "Idx：2");
    map.put("46", "Idx：2");
    map.put("68", "Idx：2");
    map.put("29", "Idx：7");
    map.put("150", "Idx：12");
    map.put("172", "Idx：12");
    map.put("194", "Idx：12");
    map.put("271", "Idx：12");
    System.out.println("排序01：");
    for (String key : map.keySet()) {
        System.out.print(key + " ");
    }
    
    map.put("293", "Idx：12");
    map.put("370", "Idx：12");
    map.put("392", "Idx：12");
    map.put("491", "Idx：12");
    map.put("590", "Idx：12");
    System.out.println("\n\n排序02：");
    for (String key : map.keySet()) {
        System.out.print(key + " ");
    }    
    
    map.remove("293");
    map.remove("370");
    map.remove("392");
    map.remove("491");
    map.remove("590");
    System.out.println("\n\n排序03：");
    for (String key : map.keySet()) {
        System.out.print(key + " ");
    }
    
}

排序01：
24 46 68 29 150 172 194 271 

排序02：
24 46 68 29 271 150 172 194 293 370 392 491 590 

排序03：
24 46 68 29 172 271 150 194 
Process finished with exit code 0

1. 根节点是黑色
2. 节点是红黑或者黑色
3. 所有子叶节点都是黑色(叶子是NIL节点，默认没有画出来)
4. 每个红色节点必须有两个黑色子节点(也同样说明一条链路上不能有链路的红色节点)
5. 黑高，从任一节点到齐每个叶子节点，经过的路径都包含相同数目的黑色节点