前言

关于集合中的Collection我们已经讲完了，接下来我们一起来看集合中的另一个大类：Map

Map的实现类

首先Map是一个接口，是一对键值对来存储信息的，K为key键，V为value值

HashMap

import java.util.HashMap;
import java.util.Map;

public class text1 {
    public static void main(String[] args) {
        /*
        *增：put(K key,V value)
        * 删：clear() remove(Object key)
        * 改：
        * 查：entrySet() get(Object key) keySet() size() values()
        * 判断：containsKey(Object key) containsValue(Object value) equals(Object o) isEmpty()
         */
        //创建一个Map集合，接口不可创建，需要创建他的实现类
        Map<Integer,String> map = new HashMap();
        map.put(1,"ymm");
        map.put(2,"yxc");
        map.put(5,"hh");
        map.put(5,"hh");
        map.put(3,"jj");
        System.out.println(map.size());
        System.out.println(map);

        //是否包含某个值
        System.out.println(map.containsKey(1));
        System.out.println(map.containsValue("ymm"));
    }
}

不难看出，又是无序且唯一的，是不是有点像hashset，他们俩个底层实现都是hash表实现。

像hashset和linkedhashset的关系一样，hashmap对应的linkedhashmap也可以对数据进行排序输出（加了一个链表进行维护），

LinkedHashMap实现类
特点：

• 唯一
• 有序（按照输入顺序进行输出）

import java.util.HashMap;
import java.util.LinkedHashMap;
import java.util.Map;

public class text1 {
    public static void main(String[] args) {
        LinkedHashMap<String,Integer> linkedHashMap = new LinkedHashMap<>();
        linkedHashMap.put("ymm",1);
        linkedHashMap.put("hh",2);
        linkedHashMap.put("jj",3);
        linkedHashMap.put("qq",4);
        System.out.println(linkedHashMap);
    }
}

HashMap原理：

import java.util.HashMap;
import java.util.LinkedHashMap;
import java.util.Map;

public class text1 {
    public static void main(String[] args) {
        HashMap<String,Integer> hashMap = new HashMap<>();
        hashMap.put("ymm",1);
        hashMap.put("xxx",2);
        hashMap.put("hh",4);
        hashMap.put("hh",3);
        System.out.println(hashMap);
        System.out.println(hashMap.size());
    }
}

为啥我存入4个数据但是现在size为3？

我们先弄清楚原理，再通过源码来验证。

我们在hashSet中讲过hash存储的那张图，这里涉及到hash碰撞的问题，看下图：

源码：（JDK17）

public class HashMap<K,V> extends AbstractMap<K,V>
    implements Map<K,V>, Cloneable, Serializable {

    private static final long serialVersionUID = 362498820763181265L;

K，V，就对应这我们上面的String，Integer，是在创建对象之前就已经确定的，可以看出HashMap继承于AbstractMap，AbstractMap也实现了Map类接口，算是一个多余的操作。

static final int DEFAULT_INITIAL_CAPACITY = 1 << 4; // aka 16默认的初始化长度为16
static final int MAXIMUM_CAPACITY = 1 << 30;定义了一个Max数
static final float DEFAULT_LOAD_FACTOR = 0.75f;负载因子
transient Node<K,V>[] table;初始数组
transient int size;集合中元素的个数
transient int modCount;数组扩容的边际值

那么是如何来创造出来HashMap对象的呢？

我们来看构造器：

	public HashMap() {
        this.loadFactor = DEFAULT_LOAD_FACTOR; // all other fields defaulted
    }//1. 无参构造器
    public HashMap(Map<? extends K, ? extends V> m) {
        this.loadFactor = DEFAULT_LOAD_FACTOR;
        putMapEntries(m, false);
    }//2. 带参构造器

1. 相当于它再调用本类的构造器，传入0.75
2. 带参构造器，相当于它先赋给0.75，再执行了putMapEntries方法

//putMapEntries方法
final void putMapEntries(Map<? extends K, ? extends V> m, boolean evict) {
        int s = m.size();
        if (s > 0) {
            if (table == null) { // pre-size
                float ft = ((float)s / loadFactor) + 1.0F;
                int t = ((ft < (float)MAXIMUM_CAPACITY) ?
                         (int)ft : MAXIMUM_CAPACITY);
                if (t > threshold)
                    threshold = tableSizeFor(t);
            }
            else if (s > threshold)
                resize();
            for (Map.Entry<? extends K, ? extends V> e : m.entrySet()) {
                K key = e.getKey();
                V value = e.getValue();
                putVal(hash(key), key, value, false, evict);
            }
        }
    }

put方法：

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

// 来自Node类：hashCode（）计算hash值
public final int hashCode() {
    return Objects.hashCode(key) ^ Objects.hashCode(value);
    //^
}
//传入key和value，再调用putVal
 final V putVal(int hash, K key, V value, boolean onlyIfAbsent,
                   boolean evict) {
        Node<K,V>[] tab; //Node类型的数组来存放
        Node<K,V> p; int n, i;
        if ((tab = table) == null || (n = tab.length) == 0)
        //初始化数组为空时候，取n为数组的初始长度
            n = (tab = resize()).length;
        if ((p = tab[i = (n - 1) & hash]) == null)
            tab[i] = newNode(hash, key, value, null);//传入自动计算hash值进行存放
        else {
            Node<K,V> e; K k;
            if (p.hash == hash &&
                ((k = p.key) == key || (key != null && key.equals(k))))
                e = p;
            else if (p instanceof TreeNode)
                e = ((TreeNode<K,V>)p).putTreeVal(this, tab, hash, key, value);
            else {
                for (int binCount = 0; ; ++binCount) {
                //链表操作，尾插
                    if ((e = p.next) == null) {
                        p.next = newNode(hash, key, value, null);
                        if (binCount >= TREEIFY_THRESHOLD - 1) // -1 for 1st
                            treeifyBin(tab, hash);
                        break;
                    }
                    if (e.hash == hash &&
                        ((k = e.key) == key || (key != null && key.equals(k))))
                        break;
                    p = e;
                }
            }
            if (e != null) { // existing mapping for key
            //hash碰撞了
                V oldValue = e.value;
                if (!onlyIfAbsent || oldValue == null)
                    e.value = value;
                    //只替换value，不换key
                afterNodeAccess(e);
                //after
                return oldValue;
            }
        }
        ++modCount;
        if (++size > threshold)
            resize();
        afterNodeInsertion(evict);
        return null;
    }

扩容resize

final Node<K,V>[] resize() {
        Node<K,V>[] oldTab = table;
        int oldCap = (oldTab == null) ? 0 : oldTab.length;
        int oldThr = threshold;
        int newCap, newThr = 0;
        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
        }
        else if (oldThr > 0) // initial capacity was placed in threshold
            newCap = oldThr;
        else {               // zero initial threshold signifies using defaults
            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;
        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;
                            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;
    }

问题：

1. 为什么负载因子为0.75
   如果负载因子为1，空间得到了很大的满足，但是会很容易产生链表，效率变低，hh。
   如果负载因子为0.5，碰撞概率地，扩容，空间利用变低
   所以取了中间值：0.75 就是时间和空间得到了折中！

2. 主数组为啥要为2^n
   原因：

   1. h&(length-1)等效于h%length，这个的前提就是主数组为2^n
   2. 防止hash冲突

HashSet底层
hh，写到这里我想笑，我们先创建一个hashset

import java.util.HashMap;
import java.util.HashSet;


public class text1 {
    public static void main(String[] args) {
        HashSet<Integer> hashSet = new HashSet<>();
    }
}

可以看到，hashset的底层实现就是一个hashmap，相当于这俩个差不多是同一个东西

TreeMap

唯一的，有序的，我们在上篇文章中说，treeset是一个二叉树实现，TreeMap也是二叉树，那么如何用一个二叉树节点来存放俩个值呢？

颜色是指代红黑树的颜色，后面的文章再说，数字为Key，"lili"为Value

源码

    static final class Entry<K,V> implements Map.Entry<K,V> {
        K key;
        V value;
        Entry<K,V> left;
        Entry<K,V> right;
        Entry<K,V> parent;
        boolean color = BLACK;

Entry的6个属性，对应我们上面图上的6个属性

构造器：

public TreeMap() {
        comparator = null;
    }

    /**
     * Constructs a new, empty tree map, ordered according to the given
     * comparator.  All keys inserted into the map must be <em>mutually
     * comparable</em> by the given comparator: {@code comparator.compare(k1,
     * k2)} must not throw a {@code ClassCastException} for any keys
     * {@code k1} and {@code k2} in the map.  If the user attempts to put
     * a key into the map that violates this constraint, the {@code put(Object
     * key, Object value)} call will throw a
     * {@code ClassCastException}.
     *
     * @param comparator the comparator that will be used to order this map.
     *        If {@code null}, the {@linkplain Comparable natural
     *        ordering} of the keys will be used.
     */
    public TreeMap(Comparator<? super K> comparator) {
        this.comparator = comparator;
    }

一个空构造器，一个带参构造器，我们要想传入自定义的比较类型，就只能使用带参构造器，具体实现，treeset文章中也有写到，我们这里就不具体实现了

put方法

public V put(K key, V value) {
//root为空
        Entry<K,V> t = root;
        if (t == null) {
        //第一个元素进入
            compare(key, key); // type (and possibly null) check

            root = new Entry<>(key, value, null);
            //封装对象
            size = 1;
            modCount++;
            return null;
        }
        int cmp;
        Entry<K,V> parent;
        // split comparator and comparable paths
        Comparator<? super K> cpr = comparator;//将外部比较器指定
        if (cpr != null) {
        //进行外部比较
            do {
                parent = t;
                cmp = cpr.compare(key, t.key);
                //完全符合二叉查找树
                if (cmp < 0)
          
                //小于0，给左
                    t = t.left;
                else if (cmp > 0)
                //大于0，给右
                    t = t.right;
                else
                    return t.setValue(value);
            } while (t != null);
        }
        else {
            if (key == null)
                throw new NullPointerException();
            @SuppressWarnings("unchecked")
                Comparable<? super K> k = (Comparable<? super K>) key;
            do {
                parent = t;
                cmp = k.compareTo(t.key);
                if (cmp < 0)
                    t = t.left;
                else if (cmp > 0)
                    t = t.right;
                else
                    return t.setValue(value);
            } while (t != null);
        }
        Entry<K,V> e = new Entry<>(key, value, parent);
        if (cmp < 0)
            parent.left = e;
        else
            parent.right = e;
        fixAfterInsertion(e);
        size++;
        modCount++;
        return null;
    }