jdk1.8HashMap的數據結構是數組+單鏈表+紅黑樹，當執行put方法的時候，會得到key的hashCode,然後進行一系列運算，得到存放在數組中的下標，數組存放的是一個鏈表的結點，所以，HashMap允許通過key計算得到的數組下標重複，這個時候，會把新插入進來的節點存放到鏈表頭部。下面看HashMap源碼
1.DEFAULT_INITIAL_CAPACITY表示數組初始容量，初始容量必須爲2的冪
2.MAXIMUM_CAPACITY 數組的最大容量
3.DEFAULT_LOAD_FACTOR 擴容因子，當數組存放的數據大於容量*0.75時會擴容
4.TREEIFY_THRESHOLD 當鏈表長度大於等於8的時候，鏈表轉成紅黑樹
5 UNTREEIFY_THRESHOLD 當紅黑樹下面的節點小於等於6時，轉成單鏈表
6 MIN_TREEIFY_CAPACITY 紅黑樹初始容量
/**
* The default initial capacity - MUST be a power of two.
*/
static final int DEFAULT_INITIAL_CAPACITY = 1 << 4; // aka 16
/**
* The maximum capacity, used if a higher value is implicitly specified
* by either of the constructors with arguments.
* MUST be a power of two <= 1<<30.
*/
static final int MAXIMUM_CAPACITY = 1 << 30;
/**
* The load factor used when none specified in constructor.
*/
static final float DEFAULT_LOAD_FACTOR = 0.75f;
/**
* The bin count threshold for using a tree rather than list for a
* bin. Bins are converted to trees when adding an element to a
* bin with at least this many nodes. The value must be greater
* than 2 and should be at least 8 to mesh with assumptions in
* tree removal about conversion back to plain bins upon
* shrinkage.
*/
static final int TREEIFY_THRESHOLD = 8;
/**
* The bin count threshold for untreeifying a (split) bin during a
* resize operation. Should be less than TREEIFY_THRESHOLD, and at
* most 6 to mesh with shrinkage detection under removal.
*/
static final int UNTREEIFY_THRESHOLD = 6;
/**
* The smallest table capacity for which bins may be treeified.
* (Otherwise the table is resized if too many nodes in a bin.)
* Should be at least 4 * TREEIFY_THRESHOLD to avoid conflicts
* between resizing and treeification thresholds.
*/
static final int MIN_TREEIFY_CAPACITY = 64;
6 Node<K,V> 單鏈表
/**
* Basic hash bin node, used for most entries. (See below for
* TreeNode subclass, and in LinkedHashMap for its Entry subclass.)
*/
static class Node<K,V> implements Map.Entry<K,V> {
final int hash;
final K key;
V value;
Node<K,V> next;
Node(int hash, K key, V value, Node<K,V> next) {
this.hash = hash;
this.key = key;
this.value = value;
this.next = next;
}
public final K getKey() { return key; }
public final V getValue() { return value; }
public final String toString() { return key + "=" + value; }
public final int hashCode() {
return Objects.hashCode(key) ^ Objects.hashCode(value);
}
public final V setValue(V newValue) {
V oldValue = value;
value = newValue;
return oldValue;
}
public final boolean equals(Object o) {
if (o == this)
return true;
if (o instanceof Map.Entry) {
Map.Entry<?,?> e = (Map.Entry<?,?>)o;
if (Objects.equals(key, e.getKey()) &&
Objects.equals(value, e.getValue()))
return true;
}
return false;
}
}
7 hash()， 得到kehash，然後把高16位與低16位做亦或運算
/**
* 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) {
int h;
return (key == null) ? 0 : (h = key.hashCode()) ^ (h >>> 16);
}
8 Node<K,V>[] table 定義一個存放鏈表節點的數組
/**
* The table, initialized on first use, and resized as
* necessary. When allocated, length is always a power of two.
* (We also tolerate length zero in some operations to allow
* bootstrapping mechanics that are currently not needed.)
*/
transient Node<K,V>[] table;
9 size HashMap存放的總數
/**
* The number of key-value mappings contained in this map.
*/
transient int size;
10 get方法，調用了getNode()
/**
* Returns the value to which the specified key is mapped,
* or {@code null} if this map contains no mapping for the key.
*
* <p>More formally, if this map contains a mapping from a key
* {@code k} to a value {@code v} such that {@code (key==null ? k==null :
* key.equals(k))}, then this method returns {@code v}; otherwise
* it returns {@code null}. (There can be at most one such mapping.)
*
* <p>A return value of {@code null} does not <i>necessarily</i>
* indicate that the map contains no mapping for the key; it's also
* possible that the map explicitly maps the key to {@code null}.
* The {@link #containsKey containsKey} operation may be used to
* distinguish these two cases.
*
* @see #put(Object, Object)
*/
public V get(Object key) {
Node<K,V> e;
return (e = getNode(hash(key), key)) == null ? null : e.value;
}
11 getNode() 方法要判斷key相等 && key.hash()相等
首先判斷 數組不爲空 && 數組下標爲 key.hash() 的數據不爲空 ，如果不滿足這些條件，返回null,
如果滿足，判斷是不是鏈表頭節點，滿足返回，不滿足就循環鏈表
/**
* Implements Map.get and related methods
*
* @param hash hash for key
* @param key the key
* @return the node, or null if none
*/
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) {
if (first.hash == hash && // always check first node
((k = first.key) == key || (key != null && key.equals(k))))
return first;
if ((e = first.next) != null) {
if (first instanceof TreeNode) //如果是紅黑樹，則走紅黑樹邏輯查找
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;
}
12 put 方法調用了putVal()
/**
* 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);
}
13
/**
* 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; //下面把tab指向了table數組
Node<K,V> p; //插入進來的key.hash 對應的數組下標的鏈表的第一個節點
int n, i; // n 數組的長度
if ((tab = table) == null || (n = tab.length) == 0)
n = (tab = resize()).length; //如果數組爲空，調用 resize()初始化數組
if ((p = tab[i = (n - 1) & hash]) == null)
//如果數組下標位置沒有存放鏈表Node,則new一個Node
tab[i] = newNode(hash, key, value, null);
else {
Node<K,V> e; K k;
if (p.hash == hash &&
((k = p.key) == key || (key != null && key.equals(k))))
e = p;//如果鏈表key的hash相等&& key的值相等，說明需要替換value
else if (p instanceof TreeNode)
//如果是紅黑樹，則使用紅黑樹的邏輯插入
e = ((TreeNode<K,V>)p).putTreeVal(this, tab, hash, key, value);
else {
//如果是鏈表，則循環鏈表，把需要插入的key,value插入到p下面，然後判斷是否需要
//鏈表轉成紅黑樹
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;
}
}
//如果e不爲空，說明key重複，需要替換value
if (e != null) { // existing mapping for key
V oldValue = e.value;
if (!onlyIfAbsent || oldValue == null)
e.value = value;
afterNodeAccess(e);
return oldValue;
}
}
++modCount;
//如果 數組的容量> 16*.75 ，擴容
if (++size > threshold)
resize();
afterNodeInsertion(evict);
return null;
}