一、put方法的流程图

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二、put方法的执行步骤

1. 首先，根据key值计算哈希值。
2. 然后判断table数组是否为空或者数组长度是否为0，是的话则要扩容，resize（）。
3. 接着，根据哈希值计算数组下标。
4. 如果这个下标位置为空，添加元素即可。
5. 如果这个下标位置不为空，则判断此下标位置的首元素的key值与要添加元素的key值是否相同，相同的话，就直接覆盖元素。
6. 不相同的话，则判断首元素是否为树节点，是的话，则向这个红黑树中插入元素（在红黑树中插入一个新节点时，首先会遍历树以找到这个新节点应该插入的位置。在这个遍历过程中，也会检查是否已经存在相同key值的节点。如果存在，也会覆盖元素；不存在，则新节点会被插入到正确的位置）。
7. 如果是链表的话，那就需要遍历链表，看key是否已经存在，存在则覆盖元素，不存在则在链表尾部插入元素。插入之后，如果链表长度大于等于8，则需要把链表转换为红黑树。
8. 最后，待所有元素处理完之后，还要判断容量是否超过阈值，超过了则需要扩容。

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三、对应源码：

       /**
     * Computes key.hashCode() and spreads (XORs) higher bits of hash
     * to lower.  Because the table uses power-of-two masking, sets of
     * hashes that vary only in bits above the current mask will
     * always collide. (Among known examples are sets of Float keys
     * holding consecutive whole numbers in small tables.)  So we
     * apply a transform that spreads the impact of higher bits
     * downward. There is a tradeoff between speed, utility, and
     * quality of bit-spreading. Because many common sets of hashes
     * are already reasonably distributed (so don't benefit from
     * spreading), and because we use trees to handle large sets of
     * collisions in bins, we just XOR some shifted bits in the
     * cheapest possible way to reduce systematic lossage, as well as
     * to incorporate impact of the highest bits that would otherwise
     * never be used in index calculations because of table bounds.
     */
    static final int hash(Object key) {//1.首先，根据key值计算哈希值。
        int h;
        return (key == null) ? 0 : (h = key.hashCode()) ^ (h >>> 16);
    }


 /**
     * Associates the specified value with the specified key in this map.
     * If the map previously contained a mapping for the key, the old
     * value is replaced.
     *
     * @param key key with which the specified value is to be associated
     * @param value value to be associated with the specified key
     * @return the previous value associated with <tt>key</tt>, or
     *         <tt>null</tt> if there was no mapping for <tt>key</tt>.
     *         (A <tt>null</tt> return can also indicate that the map
     *         previously associated <tt>null</tt> with <tt>key</tt>.)
     */
    public V put(K key, V value) {
        return putVal(hash(key), key, value, false, true);
                         //hash值
    }

    /**
     * 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
     * @param evict if false, the table is in creation mode.
     * @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;

//2.然后判断table数组是否为空或者数组长度是否为0，是的话则要扩容，resize（）。
        if ((tab = table) == null || (n = tab.length) == 0)
            n = (tab = resize()).length;

//3.接着，根据哈希值计算数组下标。
//4.如果这个下标位置为空，添加元素即可。
        if ((p = tab[i = (n - 1) & hash]) == null)
            tab[i] = newNode(hash, key, value, null);
        else {

//5.如果这个下标位置不为空，则判断此下标位置的首元素的key值与要添加元素的key值是否相同，相同的话，就直接覆盖元素。
            Node<K,V> e; K k;
            if (p.hash == hash &&
                ((k = p.key) == key || (key != null && key.equals(k))))
                e = p;

//6.不相同的话，则判断首元素是否为树节点，是的话，则向这个红黑树中插入元素
            else if (p instanceof TreeNode)
                e = ((TreeNode<K,V>)p).putTreeVal(this, tab, hash, key, value);

//7.如果是链表的话，那就需要遍历链表，看key是否已经存在，存在则覆盖元素，不存在则在链表尾部插入元素。不存在则在链表尾部插入元素。插入之后，如果链表长度大于等于8，则需要把链表转换为红黑树。
            else {
                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);//转为红黑树
                        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;

//8.最后，待所有元素处理完之后，还要判断容量是否超过阈值，超过了则需要扩容。
        if (++size > threshold)
            resize();
        afterNodeInsertion(evict);
        return null;
    }

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