几个重要的变量：

static final int DEFAULT_INITIAL_CAPACITY = 1 << 4;//初始大小

static final int MAXIMUM_CAPACITY = 1 << 30;//map中最多可以容纳的数量

static final float DEFAULT_LOAD_FACTOR = 0.75f; //与当前容量大小决定扩容极限值，如默认的容量（16），默认的扩容极限值（16 * 0.75 = 12）；如果size达到12时，进行下次扩容，容量和极限值均扩容成原来的两倍，32,24；

构造方法

public HashMap(int initialCapacity, float loadFactor) {
        if (initialCapacity < 0)
            throw new IllegalArgumentException("Illegal initial capacity: " +
                                               initialCapacity);
        if (initialCapacity > MAXIMUM_CAPACITY)
            initialCapacity = MAXIMUM_CAPACITY;
        if (loadFactor <= 0 || Float.isNaN(loadFactor))
            throw new IllegalArgumentException("Illegal load factor: " +
                                               loadFactor);
        this.loadFactor = loadFactor;
        this.threshold = tableSizeFor(initialCapacity);
    }

public HashMap(int initialCapacity) {
        this(initialCapacity, DEFAULT_LOAD_FACTOR);
    }

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

put(K key, V value)

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

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


transient Node<K,V>[] table;    //Node是单向链表的节点，When allocated, length is always a power of two.
// @return previous value, or null if none
    final V putVal(int hash, K key, V value, boolean onlyIfAbsent,
                   boolean evict) {
        Node<K,V>[] tab; Node<K,V> p; int n, i;
        if ((tab = table) == null || (n = tab.length) == 0)
            n = (tab = resize()).length;    //resize(): Initializes or doubles table size.此处时初始化
        if ((p = tab[i = (n - 1) & hash]) == null)
            tab[i] = newNode(hash, key, value, null);   //如果目标index目前的元素是空的则直接赋值就行
        else {
            Node<K,V> e; K k;
            if (p.hash == hash &&
                ((k = p.key) == key || (key != null && key.equals(k))))
                e = p;  //key已存在map中，更新值
            else if (p instanceof TreeNode)
                e = ((TreeNode<K,V>)p).putTreeVal(this, tab, hash, key, value);
            else {
                for (int binCount = 0; ; ++binCount) {
                    if ((e = p.next) == null) {
                        p.next = newNode(hash, key, value, null);   //以链表的形式将同一index的数据组织起来
                        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
                V oldValue = e.value;
                if (!onlyIfAbsent || oldValue == null)
                    e.value = value;
                afterNodeAccess(e);
                return oldValue;
            }
        }
        ++modCount;
        if (++size > threshold)
            resize();
        afterNodeInsertion(evict);
        return null;
    }


final Node<K, V>[] resize() {
   ...
     //初始大小为static final int DEFAULT_INITIAL_CAPACITY = 1 << 4;
     //计算得出新数组的大小，成倍增长 newThr = oldThr << 1
     //如果达到MAXIMUM_CAPACITY，其他不变，threshold = Integer.MAX_VALUE，return原数组
     Node<K,V>[] newTab = (Node<K,V>[])new Node[newCap];
        table = newTab; //更新成员变量的值
   ...
    //将旧数组的数据按序迁移至新数组
     
}

put方法中含有的信息较多，可以看到HashMap内部是数组+链表的形式来存储数据的，每个数据被组装成链表的节点。

get(Object key)

public V get(Object key) {
        Node<K,V> e;
        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;
        if ((tab = table) != null && (n = tab.length) > 0 &&
            (first = tab[(n - 1) & hash]) != null) {    //取到数组中对应index的链表的首节点
            if (first.hash == hash && // 从首节点开始遍历取出目标元素
                ((k = first.key) == key || (key != null && key.equals(k))))
                return first;
            if ((e = first.next) != null) {
                if (first instanceof TreeNode)
                    return ((TreeNode<K,V>)first).getTreeNode(hash, key);
                do {
                    if (e.hash == hash &&
                        ((k = e.key) == key || (key != null && key.equals(k))))
                        return e;
                } while ((e = e.next) != null);
            }
        }
        return null;
    }

如果判断元素是否是目标元素：node.hash == hash && node.key == key || (key != null && key.equals(k))

remove(Object key)

public boolean remove(Object key, Object value) {
        return removeNode(hash(key), key, value, true, true) != null;
    }
//return the node, or null if none
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;
        if ((tab = table) != null && (n = tab.length) > 0 &&
            (p = tab[index = (n - 1) & hash]) != null) {    //通过hash快速找到对应的index
            Node<K,V> node = null, e; K k; V v;
            if (p.hash == hash &&
                ((k = p.key) == key || (key != null && key.equals(k))))
                node = p;
            else if ((e = p.next) != null) {
                if (p instanceof TreeNode)
                    node = ((TreeNode<K,V>)p).getTreeNode(hash, key);
                else {
                    do {
                        if (e.hash == hash &&
                            ((k = e.key) == key ||
                             (key != null && key.equals(k)))) {
                            node = e;
                            break;
                        }
                        p = e;
                    } while ((e = e.next) != null);
                }
            }   //遍历index中的链表元素，找到目标节点
            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;
            }//将目标节点从链表中移除
        }
        return null;
    }

replace(K key, V value)

public V replace(K key, V value) {
        Node<K,V> e;
        if ((e = getNode(hash(key), key)) != null) {
            V oldValue = e.value;
            e.value = value;
            afterNodeAccess(e);
            return oldValue;
        }
        return null;
    }

//well，主要分析下面这个方法
public boolean replace(K key, V oldValue, V newValue) {
        Node<K,V> e; V v;
            //getNode()在get()方法中分析过了
        if ((e = getNode(hash(key), key)) != null &&
            ((v = e.value) == oldValue || (v != null && v.equals(oldValue)))) {
            e.value = newValue; //替换目标节点的值
            afterNodeAccess(e);
            return true;
        }
        return false;
    }

clear()

public void clear() {
        Node<K,V>[] tab;
        modCount++;
        if ((tab = table) != null && size > 0) {
            size = 0;
            for (int i = 0; i < tab.length; ++i)
                tab[i] = null;  //清空数组即可
        }
    }

Q：为什么扩容因子是0.75？