源码分析——HashMap的put,resize,containskey方法原理分析
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HashMap的put方法
/**
* 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);
1.map将指定的value和key关联起来。
2.如果map之前包含key的映射，新的值会替换旧的值。
3.map支持null的key，null的value。
put方法里面，
第一步：对key值进行hash算法操作。
/**
* 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);
设计者设计hash算法的原因：1.权衡速度，效用，质量之间的均衡考虑。 2.目前的hash集合已经分配均匀，如果碰撞频繁可以采用tree来处理。 3.高16位和低16位进行异或处理，系统损耗最小。计算细节如下：
第二步，分配容器，建立key和value的绑定关系
/**
* 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为map的Map.Entry<K,V> 接口实现，有hash值,key值,value值,Node<K,V>四个参数构成
Node<K,V>[] tab; Node<K,V> p; int n, i;
//如果table为null,则创建
if ((tab = table) == null || (n = tab.length) == 0)
n = (tab = resize()).length;
//计算index位置
if ((p = tab[i = (n - 1) & hash]) == null)
tab[i] = newNode(hash, key, value, null);
else
Node<K,V> e; K k;
//节点存在
if (p.hash == hash &&
((k = p.key) == key || (key != null && key.equals(k))))
e = p;
//结构为treeNode
else if (p instanceof TreeNode)
e = ((TreeNode<K,V>)p).putTreeVal(this, tab, hash, key, value);
//结构为链表
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;
if (e != null)
V oldValue = e.value;
if (!onlyIfAbsent || oldValue == null)
e.value = value;
afterNodeAccess(e);
return oldValue;
++modCount;
//bucket超过负载因子*0.75，进行resize()操作
if (++size > threshold)
resize();
afterNodeInsertion(evict);
return null;
判断新元素和已有元素，如果没碰撞直接放到bucket里；如果碰撞了，以链表的形式存在buckets后；如果碰撞导致链表过长(大于等于TREEIFY_THRESHOLD)，就把链表转换成红黑树；如果节点已经存在就替换old value(保证key的唯一性)；如果bucket满了(超过load factor*current capacity)，就要resize。
HashMap的resize方法
/**
* Initializes or doubles table size. If null, allocates in
* accord with initial capacity target held in field threshold.
* Otherwise, because we are using power-of-two expansion, the
* elements from each bin must either stay at same index, or move
* with a power of two offset in the new table.
*
* @return the table
*/
final Node<K,V>[] resize()
Node<K,V>[] oldTab = table;
int oldCap = (oldTab == null) ? 0 : oldTab.length;
int oldThr = threshold;
int newCap, newThr = 0;
//节点存在，旧容量>0
if (oldCap > 0)
//旧的容量>=最大默认容量，把最大容量赋值给最大默认容量
if (oldCap >= MAXIMUM_CAPACITY)
threshold = Integer.MAX_VALUE;
return oldTab;
//旧容量大小在初始容量和最大容量之间，则容量扩增为原来的2倍+1
else if ((newCap = oldCap << 1) < MAXIMUM_CAPACITY &&
oldCap >= DEFAULT_INITIAL_CAPACITY)
newThr = oldThr << 1; // double threshold
//旧threshold>0
else if (oldThr > 0) // initial capacity was placed in threshold
newCap = oldThr;
//节点不存在，capacity和threshold赋默认值
else // zero initial threshold signifies using defaults
newCap = DEFAULT_INITIAL_CAPACITY;
newThr = (int)(DEFAULT_LOAD_FACTOR * DEFAULT_INITIAL_CAPACITY);
//计算resize值的上限
if (newThr == 0)
float ft = (float)newCap * loadFactor;
newThr = (newCap < MAXIMUM_CAPACITY && ft < (float)MAXIMUM_CAPACITY ?
(int)ft : Integer.MAX_VALUE);
threshold = newThr;
@SuppressWarnings("rawtypes","unchecked")
Node<K,V>[] newTab = (Node<K,V>[])new Node[newCap];
table = newTab;
if (oldTab != null)
//把每个bucket移动到新的bucket中
for (int j = 0; j < oldCap; ++j)
Node<K,V> e;
if ((e = oldTab[j]) != null)
oldTab[j] = null;
if (e.next == null)
newTab[e.hash & (newCap - 1)] = e;
else if (e instanceof TreeNode)
((TreeNode<K,V>)e).split(this, newTab, j, oldCap);
else
Node<K,V> loHead = null, loTail = null;
Node<K,V> hiHead = null, hiTail = null;
Node<K,V> next;
do
next = e.next;
//原索引的值
if ((e.hash & oldCap) == 0)
if (loTail == null)
loHead = e;
else
loTail.next = e;
loTail = e;
//原索引+oldCap位置的值
else
if (hiTail == null)
hiHead = e;
else
hiTail.next = e;
hiTail = e;
while ((e = next) != null);
//原索引放到bucket里面
if (loTail != null)
loTail.next = null;
newTab[j] = loHead;
//原索引+oldCap位置放到bucket里面
if (hiTail != null)
hiTail.next = null;
newTab[j + oldCap] = hiHead;
return newTab;
1.如果 旧容量>最大默认容量，则旧容量=默认容量；
2.如果 初始默认容量<旧容量<最大默认容量，则旧容量=旧容量<<1;
对于原来的旧数据，如果hash值&oldCap==0，则不需要改变；如果不等于0，则需要移动到 原位置+oldCap；
例如：
假设原key的hash值二进制为 0100110 ,
map初始容量16, 二进制为 0010000 , &操作结果为0，证明原来后5位小于等于15，这些元素是不需要移动位置的。如果&操作为16，则说明旧数据可以移动到16+原位置。
3.如果 旧容量==0，则默认容量为16，threshold=12；
HashMap的containsKey方法
containsKey的核心实现跟get一样，都是去HashMap中查找是否有这个值，只是containsKey返回Boolean类型，get返回value值。
/**
* Returns <tt>true</tt> if this map contains a mapping for the
* specified key.
*
* @param key The key whose presence in this map is to be tested
* @return <tt>true</tt> if this map contains a mapping for the specified
* key.
*/
public boolean containsKey(Object key)
return getNode(hash(key), key) != null;
/**
* 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;
//判断hashcode跟bucket做与运算，是否有值
if ((tab = table) != null && (n = tab.length) > 0 &&
(first = tab[(n - 1) & hash]) != null)
//如果key的hash值跟第一个元素的hash值相等，key值相等,返回这个元素
if (first.hash == hash && // always check first node
((k = first.key) == key || (key != null && key.equals(k))))
return first;
//如果有后继节点
if ((e = first.next) != null)
//通过树查找
if (first instanceof TreeNode)
return ((TreeNode<K,V>)first).getTreeNode(hash, key);
//通过链表查找
do
if (e.hash == hash &&
((k = e.key) == key || (key != null && key.equals(k))))
return e;
while ((e = e.next) != null);
return null;
1.key的hash值跟原始容量-1做&运算，确定桶的位置
2.如果直接命中，则返回，时间复杂度为O(1)；否则，如果是树，进行树节点的查找，时间复杂度为O(logn)；如果是链表，进行链表的循环查找,时间复杂度为O(n)。
参考资料：
JDK1.8HashMap源码
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