PUT:

1. First hash the hashCode() of the key, and then calculate the index.

2. If there is no collision, put it directly in the bucket.

3. If there is a collision, there are buckets in the form of linked lists.

4. If the list is too long due to collision, convert the list to a red black tree.

5. If the node already exists, replace the old value (to ensure the uniqueness of the key).

6. If the bucket is full (more than load factor*current capacity), resize it.

public V put(K key, V value) {
    // hash the hashCode() of the key
    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;
    // Create if tab is empty
    if ((tab = table) == null || (n = tab.length) == 0)
        n = (tab = resize()).length;
    // Calculate index and handle null
    if ((p = tab[i = (n - 1) & hash]) == null)
        tab[i] = newNode(hash, key, value, null);
    else {
        Node<K,V> e; K k;
        // Node existence
        if (p.hash == hash &&
            ((k = p.key) == key || (key != null && key.equals(k))))
            e = p;
        // The chain is a tree.
        else if (p instanceof TreeNode)
            e = ((TreeNode<K,V>)p).putTreeVal(this, tab, hash, key, value);
        // The chain is a list
        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);
                    break;
                }
                if (e.hash == hash &&
                    ((k = e.key) == key || (key != null && key.equals(k))))
                    break;
                p = e;
            }
        }
        // Write in
        if (e != null) { // existing mapping for key
            V oldValue = e.value;
            if (!onlyIfAbsent || oldValue == null)
                e.value = value;
            afterNodeAccess(e);
            return oldValue;
        }
    }
    ++modCount;
    // load factor*current capacity, resize exceeded
    if (++size > threshold)
        resize();
    afterNodeInsertion(evict);
    return null;
}

GET:

1. The first node in the bucket, directly hit;

2. If there is a conflict, use key.equals(k) to find the corresponding entry

3. If it is a tree, search through key.equals(k) in the tree, O(logn);

4. If it is a linked list, search through key.equals(k) in the linked list, O(n).

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) {
        // Direct hit
        if (first.hash == hash && // always check first node
            ((k = first.key) == key || (key != null && key.equals(k))))
            return first;
        // Missed hit
        if ((e = first.next) != null) {
            // get in trees
            if (first instanceof TreeNode)
                return ((TreeNode<K,V>)first).getTreeNode(hash, key);
            // get in the linked list
            do {
                if (e.hash == hash &&
                    ((k = e.key) == key || (key != null && key.equals(k))))
                    return e;
            } while ((e = e.next) != null);
        }
    }
    return null;
}