Android 开发也要懂得数据结构 - HashMap源码
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文章目录
- 1.HashMap特点
- 2.HashMap 的继承关系
- 3.HashMap常用方法
- 3.1 构造方法
- 3.2 放入元素 put(K key, V value)
- 3.3 HashMap核心 putVal(int hash, K key, V value, boolean onlyIfAbsent,boolean evict)
- 3.4 扩容 resize()
- 3.5 查找元素 get(Object key)
- 3.6 移除元素 remove(Object key)
- 3.7 清空 clear()
- 3.8 长度 size()
- 3.9 获取所有元素的集合 entrySet()
- 3.10 获取key的集合 keySet()
- 3.11 是否包含key containsKey(Object key)
- 3.12 是否包含value containsValue(Object value)
- HashMap不仅是android开发中常用的数据结构，面试也是高频出现，所以了解一下源码还是非常必要的。
- 本文章使用的是 JDK1.8 ，不同版本源码有差异。
- 文章里面的图片来自 极客时间，王争老师的数据结构与算法课。
- 极客时间 - 数据结构与算法
1.HashMap特点
- Collection 是集合，有数组（ArrayList）查找快增删慢，有链表（LinkList）增删快查找慢，Map 就是数组与链表的结合体，结合了两的优点。
- HashMap 的数据关系是 key 到 value 的映射关系，key 是唯一的，value 是可以重复的。
- HashMap 的 Hash , 是因为 key 是需要计算哈希值，这种数组就是散列表。
- HashMap 可以理解为 key计算后的位置用 数组 保存，数组里面的内容放着 链表 ，链表的节点是 key-value 一个个保存起来，查找的时候，快速找到数组中对应的位置，然后遍历链表。
- HashMap 非线程安全，可以用 HashTable。
- HashMap 数据是无序的，需要有序的使用 LinkedHashMap。
2.HashMap 的继承关系
- HashMap继承于 Map,而LinkedHashMap 是继承于HashMap。
3.HashMap常用方法
3.1 构造方法
- 默认构造方法，只初始化了扩容因数，0.75，就是HashMap的数组容量使用了75%，就要进行扩容操作了，注释还说明初始容量为16。
/**
* Constructs an empty <tt>HashMap</tt> with the default initial capacity
* (16) and the default load factor (0.75).
*/
public HashMap()
this.loadFactor = DEFAULT_LOAD_FACTOR; // all other fields defaulted
- 自定义初始容量和扩容因子的构造方法，initialCapacity为初始容量，最大不超过2的30次方，loadFactor为扩容因子，必需为大于0的浮点数。
/**
* 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;
/**
* Constructs an empty <tt>HashMap</tt> with the specified initial
* capacity and load factor.
*
* @param initialCapacity the initial capacity
* @param loadFactor the load factor
* @throws IllegalArgumentException if the initial capacity is negative
* or the load factor is nonpositive
*/
public HashMap(int initialCapacity, float loadFactor)
//小于0走异常处理。
if (initialCapacity < 0)
throw new IllegalArgumentException("Illegal initial capacity: " +
initialCapacity);
//最大不能超过1左移30位，也就是2的30次方，非常大的数。
if (initialCapacity > MAXIMUM_CAPACITY)
initialCapacity = MAXIMUM_CAPACITY;
//如果扩容因子小于0，或者不是浮点数，报异常处理。
if (loadFactor <= 0 || Float.isNaN(loadFactor))
throw new IllegalArgumentException("Illegal load factor: " +
loadFactor);
this.loadFactor = loadFactor;
this.threshold = tableSizeFor(initialCapacity);
- 传入Map的构造方法，默认扩容因子也是0.75，可以将Map转化为HashMap。
/**
* Constructs a new <tt>HashMap</tt> with the same mappings as the
* specified <tt>Map</tt>. The <tt>HashMap</tt> is created with
* default load factor (0.75) and an initial capacity sufficient to
* hold the mappings in the specified <tt>Map</tt>.
*
* @param m the map whose mappings are to be placed in this map
* @throws NullPointerException if the specified map is null
*/
public HashMap(Map<? extends K, ? extends V> m)
this.loadFactor = DEFAULT_LOAD_FACTOR;
putMapEntries(m, false);
3.2 放入元素 put(K key, V value)
- 首先看 hash(Object key) 方法，就是判断元素存放在数组的位置，如果空就返回 0，否则 key 的哈希值用临时变量 h 保存，再和 h 无符号右移16位的结果（>>>是无符号右移，高位补0），做异或操作（^是异或），算出存放在数组的位置，这种哈希计算过的数组其实就是 散列表。如果计算出来的结果一样，也就是哈希碰撞，那数据后面会用链表存放。
- 下图就是散列表。
- putVal(hash(key), key, value, false, true)，方法有五个参数，第一个是计算的位置，第二个是 key ， 第三个是value ， 第四个是否修改已存在的值，最后一个参数看意思是创建表格。
- 如果key 为空，null 是无法计算哈希值的，就返回0，所以 HashMap 是可以放一个 key 为空的元素的。
/**
* 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);
/**
* 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;
//如果key为空，就返回0
//否者key的哈希码 异或 key无符号右移16位的结果
return (key == null) ? 0 : (h = key.hashCode()) ^ (h >>> 16);
- Node<K,V> 节点类，我们可以看到，是单链表结构，还重写了equals(Object o)。
/**
* 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;
//单链表结构
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;
//重写equals
public final boolean equals(Object o)
if (o == this)
return true;
if (o instanceof Map.Entry)
Map.Entry<?,?> e = (Map.Entry<?,?>)o;
//对比key和value
if (Objects.equals(key, e.getKey()) &&
Objects.equals(value, e.getValue()))
return true;
return false;
3.3 HashMap核心 putVal(int hash, K key, V value, boolean onlyIfAbsent,boolean evict)
-
这个方法看到Node<K,V>[]，HashMap的结构就能理解了吧。
-
HashMap为了解决散列冲突，就用了链表法。
-
存放数据的链表，在长度在 8以下 的时候，是 链表 存储，8以上 或者数组长度大于64时就红黑树存储。
-
由于方法细节太多，直接写注释一步步好理解。
/**
* 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;
/**
* 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)
//分别是散列表，节点，散列表长度，索引位置
Node<K,V>[] tab; Node<K,V> p; int n, i;
//如果散列表为空或者散列表长度为0
if ((tab = table) == null || (n = tab.length) == 0)
//resize()是创建哈希表，长度为16，并且将长度赋值给n
n = (tab = resize()).length;
//找到hash值在当前哈希表中的位置，该位置的节点赋值给p，且判断该位置是否为空
if ((p = tab[i = (n - 1) & hash]) == null)
//如果为空就把这个元素放在这个位置
tab[i] = newNode(hash, key, value, null);
else
Node<K,V> e; K k;
//如果hash值相同，key也相同
if (p.hash == hash &&
((k = p.key) == key || (key != null && key.equals(k))))
//将节点p赋值给e
e = p;
//如果p是树节点
else if (p instanceof TreeNode)
//创建一个树节点赋值给e
e = ((TreeNode<K,V>)p).putTreeVal(this, tab, hash, key, value);
else
//链表节点，就遍历链表
for (int binCount = 0; ; ++binCount)
//将p的下一节点赋值给e，且为空
if ((e = p.next) == null)
//找到链表尾部，插入新的节点
p.next = newNode(hash, key, value, null);
//如果链表的长度大于8的时候
if (binCount >= TREEIFY_THRESHOLD - 1) // -1 for 1st
//链表转树结构
treeifyBin(tab, hash);
break;
//遍历到的位置已经有元素了，这里就是遍历链表一直循环next，直到为空停止
if (e.hash == hash &&
((k = e.key) == key || (key != null && key.equals(k))))
break;
p = e;
//如果当前节点不为空，前面的操作除了最后一个else,其他就是找到已存在的节点
if (e != null) // existing mapping for key
//当前节点的值赋值给oldValue
V oldValue = e.value;
//如果不修改值，或者oldValue 为空（存储value为空）
if (!onlyIfAbsent || oldValue == null)
//就修改当前节点的值
e.value = value;
afterNodeAccess(e);
//修改值，在这return,不再增加数据的长度
return oldValue;
++modCount;
//添加好元素，长度+1
if (++size > threshold)
//扩容操作
resize();
afterNodeInsertion(evict);
return null;
/**
* Replaces all linked nodes in bin at index for given hash unless
* table is too small, in which case resizes instead.
*/
final void treeifyBin(Node<K,V>[] tab, int hash)
int n, index; Node<K,V> e;
//如果哈希表为空，或者哈希表长度小于64，优先扩容，而不是转为红黑树
if (tab == null || (n = tab.length) < MIN_TREEIFY_CAPACITY)
//扩容
resize();
//否则判断不为空就转为树节点
else if ((e = tab[index = (n - 1) & hash]) != null)
TreeNode<K,V> hd = null, tl = null;
do
TreeNode<K,V> p = replacementTreeNode(e, null);
if (tl == null)
hd = p;
else
p.prev = tl;
tl.next = p;
tl = p;
while ((e = e.next) != null);
if ((tab[index] = hd) != null)
hd.treeify(tab);
3.4 扩容 resize()
- 扩容为原来的2倍大小，扩容完，需要重写计算位置，重写排位置。
/**
* 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;
if (oldCap > 0)
//最大不能超过Integer.MAX_VALUE
if (oldCap >= MAXIMUM_CAPACITY)
threshold = Integer.MAX_VALUE;
return oldTab;
//扩容为原来的2倍
else if ((newCap = oldCap << 1) < MAXIMUM_CAPACITY &&
oldCap >= DEFAULT_INITIAL_CAPACITY)
newThr = oldThr << 1; // double threshold
else if (oldThr > 0) // initial capacity was placed in threshold
newCap = oldThr;
else // zero initial threshold signifies using defaults
//创建默认大小，长度16
newCap = DEFAULT_INITIAL_CAPACITY;
//阈值0.75 * 16
newThr = (int)(DEFAULT_LOAD_FACTOR * DEFAULT_INITIAL_CAPACITY);
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)
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 // preserve order
Node<K,V> loHead = null, loTail = null;
Node<K,V> hiHead = null, hiTail = null;
Node<K,V> next;
do
next = e.next;
if ((e.hash & oldCap) == 0)
if (loTail == null)
loHead = e;
else
loTail.next = e;
loTail = e;
else
if (hiTail == null)
hiHead = e;
else
hiTail.next = e;
hiTail = e;
while ((e = next) != null);
if (loTail != null)
loTail.next = null;
newTab[j] = loHead;
if (hiTail != null)
hiTail.next = null;
newTab[j + oldCap] = hiHead;
return newTab;
3.5 查找元素 get(Object key)
- 查找比较简单，key为空，就是hash为0，有就返回，没有就返回null。
- key 不为空，找到就返回 value，找不到就返回null。
/**
* 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)
*/
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