前言
面試一般都會問到HashMap
整理內容自慕課網
https://coding.imooc.com/lesson/337.html#mid=24128
參考博客:https://www.cnblogs.com/aspirant/p/8906018.html
內容講的是HashTable,裏面數據結構和如何擴容說得挺好
內容
- 圖解數據結構
- 圖解說明數據結構基礎
- 圖解說明HashMap原理
- 源碼分析
- 源碼分析HashMap的創建以及擴容
- 節點的插入、查詢與刪除
- 如何提高散列以及衝突的解決辦法
圖解說明HashMap
HashMap是如何實現快速索引的?
數組
數組的本質是一塊連續的內存空間,因爲內存空間連續,我們只要知道首地址的位置,就能很快定位所有數據,所以數組的有點很明顯,就是通過下標可以快速尋址
缺點也是明顯的,就是對於有序的數據,如果想要插入,就要移動大量的位置
單鏈表
node {
T data;
node next;
}
插入刪除方便,查詢O(n)
HashMap
HashMap既要插入高效,又要查詢高效,顯然就是結合了數組和鏈表二者的優點
爲了快速定位,又要插入數據,如果保持插入的動作不變,那麼爲了能夠定位,我們就需要知道插入的值和數組下標之間的關係
假設數據是一個int類型的數組
pos = key % size
key就是插入的數據,size是數組的大小,pos就是下標
那麼插入和查找的時候都需要通過這個求模操作
但是這很顯然有一個問題
兩個插入的值100和200對一個size爲10的數組求模,得到的結果都是0,這顯然是不允許出現的情況
這就涉及到了HashMap的衝突問題
要解決這一衝突,有一個很簡單的辦法,就是在同一個下標用單鏈表來進行擴展
就上面的例子來說就是100的這個節點位置有一個指向200的指針
查找時將有一個比較值並且比較其next的過程
那麼問題就又繞回來了,單鏈表查詢低效(但其實夠用了)
在JDK1.8版本中對這一問題做了優化,將單鏈表過長的時候將轉化爲紅黑樹以提高查詢速率,這裏不對紅黑樹做講解
HashMap源碼分析
resize與hash比較
構造一個HashMap
public static void main(String[] args) {
Map<String, String> map = new HashMap<>();
map.put("100", "test");
}
查看它的構造方法
/**
* 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
}
對成員變量loadFactor負載因子賦值0.75f
再來查看HashMap的put操作
/**
* 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);
}
裏面是一個putVal
/**
* 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; Node<K,V> p; int n, i;
if ((tab = table) == null || (n = tab.length) == 0)
n = (tab = resize()).length;
if ((p = tab[i = (n - 1) & hash]) == null)
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;
else if (p instanceof TreeNode)
e = ((TreeNode<K,V>)p).putTreeVal(this, tab, hash, key, value);
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) { // existing mapping for key
V oldValue = e.value;
if (!onlyIfAbsent || oldValue == null)
e.value = value;
afterNodeAccess(e);
return oldValue;
}
}
++modCount;
if (++size > threshold)
resize();
afterNodeInsertion(evict);
return null;
}
putVal的幾件事情
- 判斷table是否爲空 進行resize
- hash判斷是直接插入還是鏈表插入
resize方法有三個重要的參數:負載因子、容量、閾值
下面是初始化時的參數
newCap = DEFAULT_INITIAL_CAPACITY; // 1 << 4 aka 16
newThr = (int)(DEFAULT_LOAD_FACTOR * DEFAULT_INITIAL_CAPACITY);
噹噹前的table大小超過閾值的時候就會擴容
if (oldCap > 0) {
if (oldCap >= MAXIMUM_CAPACITY) {
threshold = Integer.MAX_VALUE;
return oldTab;
}
else if ((newCap = oldCap << 1) < MAXIMUM_CAPACITY &&
oldCap >= DEFAULT_INITIAL_CAPACITY)
newThr = oldThr << 1; // double threshold
}
擴容是舊的容量左移一位
再看一下它的hash判斷
if ((p = tab[i = (n - 1) & hash]) == null)
n是table的大小 大小減1之後hash得到的i(這裏涉及hash的計算之後再說 hash自然是根據內容得到的)
面試問題:爲什麼初始大小爲2的倍數
- 最主要的是:2的倍數減1之後會得到後面全爲1的二進制位,hash與運算可以有多種結果,即提高散列度
- 通用的,減少內存分頁的碎片
- 計算機最擅長與運算等
如何Hash
public V put(K key, V value) {
return putVal(hash(key), key, value, false, true);
}
putVal的第一個參數就是hash,那麼如何得到這個根據key獲取到的hash值呢
static final int hash(Object key) {
int h;
return (key == null) ? 0 : (h = key.hashCode()) ^ (h >>> 16);
}
這裏邏輯很簡單
- 首先判空,如果爲空則直接返回0
- 如果不爲空則通過native方法hasCode獲取int值
- 與這個值右移十六位之後的結果進行異或運算
這裏的異或操作也是爲了提高散列度…至於爲什麼我也不知道
這是面試問題中Hash函數如何減少衝突的一個回答
衝突的解決
其實就是
if ((p = tab[i = (n - 1) & hash]) == null)
這個else裏面的內容了
這裏簡單梳理一下邏輯
if (p.hash == hash &&
((k = p.key) == key || (key != null && key.equals(k))))
e = p;
else if (p instanceof TreeNode)
e = ((TreeNode<K,V>)p).putTreeVal(this, tab, hash, key, value);
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;
}
}
這裏有三種可能
- 插入的新節點的hash與key值都與當前節點相同
- 舊節點是一個黑紅樹的節點
- 舊節點是一個單鏈表的節點
插入屬於數據結構的知識了,這裏看一下單鏈表到紅黑樹轉化的邏輯
if (binCount >= TREEIFY_THRESHOLD - 1) // -1 for 1st
treeifyBin(tab, hash);
名字非常直觀了 如果單鏈表的長度超過了樹化的閾值那麼就進行紅黑樹化
默認的樹化閾值爲8
擴容與非擴容後的內存排布
來分析一下resize
Node<K,V>[] oldTab = table;
int oldCap = (oldTab == null) ? 0 : oldTab.length;
int oldThr = threshold;
int newCap, newThr = 0;
if (oldCap > 0) {
if (oldCap >= MAXIMUM_CAPACITY) {
threshold = Integer.MAX_VALUE;
return oldTab;
}
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
newCap = DEFAULT_INITIAL_CAPACITY;
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;
前面都在進行初始化
如果oldTable是有值的,那麼就要把oldTable的節點給複製到newTable
這裏有三個判斷
- e.next == null 也就是沒有衝突的情況
- e instanceof TreeNode 衝突解決是紅黑樹的形式
- 衝突解決是單鏈表的形式
分支選擇來完成複製
- 沒有衝突
newTab[e.hash & (newCap - 1)] = e;
根據hash來確定位置 這裏的下標是newCap-1來hash的,這和putVal的hash判斷一致
2. 紅黑樹的split 略
3. 單鏈表的處理
//如果擴容後,元素的index依然與原來一樣,那麼使用這個head和tail指針
Node<K,V> loHead = null, loTail = null
//如果擴容後,元素的index=index+oldCap,那麼使用這個head和tail指針
Node<K,V> hiHead = null, hiTail = null
Node<K,V> next;
do {
next = e.next;
//這個地方直接通過hash值與oldCap進行與操作得出元素在新數組的index
if ((e.hash & oldCap) == 0) {
if (loTail == null)
loHead = e;
else
loTail.next = e;
//tail指針往後移動一位,維持順序
loTail = e;
}
else {
if (hiTail == null)
hiHead = e;
else
hiTail.next = e;
//tail指針往後移動一位,維持順序
hiTail = e;
}
} while ((e = next) != null);
if (loTail != null) {
loTail.next = null;
//還是原來的index
newTab[j] = loHead;
}
if (hiTail != null) {
hiTail.next = null;
//index = index + oldCap
newTab[j + oldCap] = hiHead;
這裏這麼做可以保證它與舊的鏈表順序相同
HashMap的查詢
public static void main(String[] args) {
Map<String, String> map = new HashMap<>();
map.put("100", "test");
map.get("100");
}
/**
* 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;
}
調用getNode方法
/**
* 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)
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;
}
這裏也沒什麼特別的,就和平時寫業務邏輯的代碼一樣判空之類操作
若都不爲空爲比較key值,若first不等則分別按照紅黑樹和單鏈表來繼續比較
HashMap刪除
/**
* Removes the mapping for the specified key from this map if present.
*
* @param key key whose mapping is to be removed from the map
* @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 remove(Object key) {
Node<K,V> e;
return (e = removeNode(hash(key), key, null, false, true)) == null ?
null : e.value;
}
看一下removeNode
/**
* Implements Map.remove and related methods
*
* @param hash hash for key
* @param key the key
* @param value the value to match if matchValue, else ignored
* @param matchValue if true only remove if value is equal
* @param movable if false do not move other nodes while removing
* @return the node, or null if none
*/
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) {
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;
}
刪除的邏輯開始和查找一樣,找到之後獲取這個node
然後分別根據無衝突、黑紅樹、單鏈表三種分支來操作刪除
共同需要做的就是–size等
HashMap的序列化
HashMap的序列化主要是調用了writeObject方法