增删改查方法
1.put,增,改
public V put(K key, V value) { return putVal(hash(key), key, value, false, true); }
public V putIfAbsent(K key, V value) { return putVal(hash(key), key, value, true, 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; if ((tab = table) == null || (n = tab.length) == 0)//空table则建table n = (tab = resize()).length;//建好table,n为table长度 if ((p = tab[i = (n - 1) & hash]) == null)//hash之后的索引位置为空则直接放进去,p为索引为i的链表头 tab[i] = newNode(hash, key, value, null); else {//不为空则插入 Node<K,V> e; K k; if (p.hash == hash && ((k = p.key) == key || (key != null && key.equals(k))))//如果链表头和要插入的节点相同 e = p; else if (p instanceof TreeNode)//如果是红黑树则采用红黑树的put e = ((TreeNode<K,V>)p).putTreeVal(this, tab, hash, key, value); else { for (int binCount = 0; ; ++binCount) {//binCount为了统计是否到了变红黑树的阀值 if ((e = p.next) == null) {//找到链表的尾节点,插入 p.next = newNode(hash, key, value, null); if (binCount >= TREEIFY_THRESHOLD - 1) //如果超过变红黑树的阀值,变红黑树 treeifyBin(tab, hash); break; } if (e.hash == hash && ((k = e.key) == key || (key != null && key.equals(k))))//如果e和要插入的节点相同 break; p = e; } } if (e != null) { //如果e不为空,根据参数决定是否替换老的值 V oldValue = e.value; if (!onlyIfAbsent || oldValue == null) e.value = value; afterNodeAccess(e);//对e进行操作后的函数,主要用于linkedHashMap实现LRU return oldValue; } } ++modCount;//结构性更改次数增加 if (++size > threshold)//如果size超过阀值,resize resize(); afterNodeInsertion(evict); return null; }
2.remove,删
public boolean remove(Object key, Object value) { return removeNode(hash(key), key, value, true, true) != 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; if ((tab = table) != null && (n = tab.length) > 0 && (p = tab[index = (n - 1) & hash]) != null) {//table不为空且table槽 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); } } 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; }
3.get
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) {//查询槽 if (first.hash == hash && // always check first node ((k = first.key) == key || (key != null && key.equals(k))))//对比首节点 return first; if ((e = first.next) != null) { if (first instanceof TreeNode) return ((TreeNode<K,V>)first).getTreeNode(hash, key);//如果是树节点,查询树节点 do {//查询链表 if (e.hash == hash && ((k = e.key) == key || (key != null && key.equals(k)))) return e; } while ((e = e.next) != null); } } return null; }