HashMap vs. TreeMap vs. Hashtable vs. LinkedHashMap
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原文:http://www.importnew.com/8658.html
Map是最重要的数据结构。这篇文章中,我会带你们看看HashMap, TreeMap, HashTable和LinkedHashMap的区别。
1. Map概览
Java SE中有四种常见的Map实现——HashMap, TreeMap, Hashtable和LinkedHashMap。如果我们使用一句话来分别概括它们的特点,就是:
- HashMap就是一张hash表,键和值都没有排序。
- TreeMap以红-黑树结构为基础,键值按顺序排列。
- LinkedHashMap保存了插入时的顺序。
- Hashtable是同步的(而HashMap是不同步的)。所以如果在线程安全的环境下应该多使用HashMap,而不是Hashtable,因为Hashtable对同步有额外的开销。
2. HashMap
如果HashMap的键(key)是自定义的对象,那么需要按规则定义它的equals()和hashCode()方法。
class Dog {String color;Dog(String c) {color = c;}public String toString() {return color + " dog";}}public class TestHashMap {public static void main(String[] args) {HashMap hashMap = new HashMap();Dog d1 = new Dog("red");Dog d2 = new Dog("black");Dog d3 = new Dog("white");Dog d4 = new Dog("white");hashMap.put(d1, 10);hashMap.put(d2, 15);hashMap.put(d3, 5);hashMap.put(d4, 20);// print sizeSystem.out.println(hashMap.size());// loop HashMapfor (Entry entry : hashMap.entrySet()) {System.out.println(entry.getKey().toString() + " - " + entry.getValue());}}}
输出
4white dog - 5black dog - 15red dog - 10white dog - 20
注意,我们错误的将”white dogs”添加了两次,但是HashMap却接受了两只”white dogs”。这不合理(因为HashMap的键不应该重复),我们会搞不清楚真正有多少白色的狗存在。
Dog类应该定义如下:
class Dog {String color;Dog(String c) {color = c;}public boolean equals(Object o) {if (this == o) {return true;}if (o instanceof Dog) {return ((Dog) o).color.equals(this.color);}return false;}public int hashCode() {return color.length();}public String toString() {return color + " dog";}}
输出:
3red dog - 10white dog - 20black dog - 15
输出结果如上是因为HashMap不允许有两个相等的元素存在。默认情况下(也就是类没有实现hashCode()和equals()方法时),会使用Object类中的这两个方法。Object类中的hashCode()对于不同的对象会返回不同的整数,而只有两个引用指向的同样的对象时 equals()才会返回true。
3. TreeMap
TreeMap的键按顺序排列。让我们先看个例子看看什么叫作“键按顺序排列”。
class Dog {String color;Dog(String c) {color = c;}public boolean equals(Object o) {if (this == o) {return true;}if (o instanceof Dog) {return ((Dog) o).color.equals(this.color);}return false;}public int hashCode() {return color.length();}public String toString() {return color + " dog";}}public class TestTreeMap {public static void main(String[] args) {Dog d1 = new Dog("red");Dog d2 = new Dog("black");Dog d3 = new Dog("white");Dog d4 = new Dog("white");TreeMap treeMap = new TreeMap();treeMap.put(d1, 10);treeMap.put(d2, 15);treeMap.put(d3, 5);treeMap.put(d4, 20);for (Entry entry : treeMap.entrySet()) {System.out.println(entry.getKey() + " - " + entry.getValue());}}}
输出:
Exception in thread "main" java.lang.ClassCastException: collection.Dog cannot be cast to java.lang.Comparable at java.util.TreeMap.put(Unknown Source) at collection.TestHashMap.main(TestHashMap.java:35)
因为TreeMap按照键的顺序进行排列对象,所以键的对象之间需要能够比较,所以就要实现Comparable接口。你可以使用String作为键,String已经实现了Comparable接口。
我们来修改下Dog类,让它实现Comparable接口。
class Dog implements Comparable<Dog> {String color;int size;Dog(String c, int s) {color = c;size = s;}public String toString() {return color + " dog";}@Overridepublic int compareTo(Dog o) {return o.size - this.size;}}public class TestTreeMap {public static void main(String[] args) {Dog d1 = new Dog("red", 30);Dog d2 = new Dog("black", 20);Dog d3 = new Dog("white", 10);Dog d4 = new Dog("white", 10);TreeMap treeMap = new TreeMap();treeMap.put(d1, 10);treeMap.put(d2, 15);treeMap.put(d3, 5);treeMap.put(d4, 20);for (Entry entry : treeMap.entrySet()) {System.out.println(entry.getKey() + " - " + entry.getValue());}}}
输出:
red dog - 10black dog - 15white dog - 20
结果根据键的排列顺序进行输出,在我们的例子中根据size排序的。
如果我们将“Dog d4 = new Dog(“white”, 10);”替换成“Dog d4 = new Dog(“white”, 40);”,那么输出会变成:
white dog - 20red dog - 10black dog - 15white dog - 5
这是因为TreeMap使用compareTo()方法来比较键值的大小,size不相等的狗是不同的狗。
4. Hashtable
Java文档写道:
HashMap类和Hashtable类几乎相同,不同之处在于HashMap是不同步的,也允许接受null键和null值。
5. LinkedHashMap
LinkedHashMap is a subclass of HashMap. That means it inherits the features of HashMap. In addition, the linked list preserves the insertion-order.
Let’s replace the HashMap with LinkedHashMap using the same code used for HashMap.
LinkedHashMap是HashMap的子类,所以LinkedHashMap继承了HashMap的一些属性,它在HashMap基础上增加的特性就是保存了插入对象的顺序。
class Dog {String color;Dog(String c) {color = c;}public boolean equals(Object o) {if (this == o) {return true;}if (o instanceof Dog) {return ((Dog) o).color.equals(this.color);}return false;}public int hashCode() {return color.length();}public String toString() {return color + " dog";}}public class TestHashMap {public static void main(String[] args) {Dog d1 = new Dog("red");Dog d2 = new Dog("black");Dog d3 = new Dog("white");Dog d4 = new Dog("white");LinkedHashMap linkedHashMap = new LinkedHashMap();linkedHashMap.put(d1, 10);linkedHashMap.put(d2, 15);linkedHashMap.put(d3, 5);linkedHashMap.put(d4, 20);for (Entry entry : linkedHashMap.entrySet()) {System.out.println(entry.getKey() + " - " + entry.getValue());}}}
输出:
red dog - 10black dog - 15white dog - 20
如果我们使用HashMap的话,输出将会如下,会打乱插入的顺序:
red dog - 10white dog - 20black dog - 15
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