为什么哈希表(java.util.HashMap
)按 long , int , byte 和 short ?
见下面的代码:
public class Main {
private static final int INITIAL_CAPACITY = 10;
public static void main(String[] args) {
final Map<Long, Long> longMap = new HashMap<>(INITIAL_CAPACITY);
final Map<Integer, Integer> integerMap = new HashMap<>(INITIAL_CAPACITY);
final Map<Byte, Byte> byteMap = new HashMap<>(INITIAL_CAPACITY);
final Map<Short, Short> shortMap = new HashMap<>(INITIAL_CAPACITY);
final Map<Double, Double> doubleMap = new HashMap<>(INITIAL_CAPACITY);
final Map<Float, Float> floatMap = new HashMap<>(INITIAL_CAPACITY);
final Map<BigDecimal, BigDecimal> bigDecimalMap = new HashMap<>(INITIAL_CAPACITY);
final Map<String, String> stringMap = new HashMap<>(INITIAL_CAPACITY);
final Random random = new Random();
for(int i=0; i < 100; i++){
int value = random.nextInt(10);
longMap.put(Long.valueOf(value), Long.valueOf(value));
integerMap.put(Integer.valueOf(value), Integer.valueOf(value));
byteMap.put(Byte.valueOf((byte)value), Byte.valueOf((byte)value));
shortMap.put(Short.valueOf((short)value), Short.valueOf((short)value));
doubleMap.put(Double.valueOf(value), Double.valueOf(value));
floatMap.put(Float.valueOf(value), Float.valueOf(value));
bigDecimalMap.put(BigDecimal.valueOf(value), BigDecimal.valueOf(value));
stringMap.put(String.valueOf(value), String.valueOf(value));
}
System.out.println("\n========== SORTED ==========\n");
System.out.println("Map<Long, Long>: " + longMap);
System.out.println("Map<Integer, Integer>: " + integerMap);
System.out.println("Map<Byte, Byte>: " + byteMap);
System.out.println("Map<Short, Short>: " + shortMap);
System.out.println("\n======== NOT SORTED ========\n");
System.out.println("Map<Double, Double>: " + doubleMap);
System.out.println("Map<Float, Float>: " + floatMap);
System.out.println("Map<BigDecimal, BigDecimal>: " + bigDecimalMap);
System.out.println("Map<String, String>: " + stringMap);
}
}
输出此程序:
========== SORTED ==========
Map<Long, Long> : {0=0, 1=1, 2=2, 3=3, 4=4, 5=5, 6=6, 7=7, 8=8, 9=9}
Map<Integer, Integer> : {0=0, 1=1, 2=2, 3=3, 4=4, 5=5, 6=6, 7=7, 8=8, 9=9}
Map<Byte, Byte> : {0=0, 1=1, 2=2, 3=3, 4=4, 5=5, 6=6, 7=7, 8=8, 9=9}
Map<Short, Short> : {0=0, 1=1, 2=2, 3=3, 4=4, 5=5, 6=6, 7=7, 8=8, 9=9}
======== NOT SORTED ========
Map<Double, Double> : {0.0=0.0, 3.0=3.0, 6.0=6.0, 7.0=7.0, 2.0=2.0, 1.0=1.0, 4.0=4.0, 9.0=9.0, 8.0=8.0, 5.0=5.0}
Map<Float, Float> : {1.0=1.0, 0.0=0.0, 4.0=4.0, 3.0=3.0, 5.0=5.0, 2.0=2.0, 8.0=8.0, 9.0=9.0, 7.0=7.0, 6.0=6.0}
Map<BigDecimal, BigDecimal> : {6=6, 0=0, 5=5, 9=9, 7=7, 8=8, 3=3, 4=4, 2=2, 1=1}
Map<String, String> : {3=3, 2=2, 1=1, 0=0, 7=7, 6=6, 5=5, 4=4, 9=9, 8=8}
答案 0 :(得分:5)
因为对于Long
,Integer
,Byte
和Short
,重写的int hashCode()
方法很简单,并返回数值本身。由于哈希码用于索引将元素放置在哈希映射中的特定桶,因此较低的哈希导致较低的索引。请注意,保留的顺序是显而易见的:通过在哈希映射中添加其他元素,可能会将更多元素放置在同一个桶中,因此您看不到保证,这只是可能产生的副作用。 HashMap
不是已排序的集合,您不应该这样使用。当你需要排序对时,TreeMap
是你要走的路。
对于Double
,Float
,BigInteger
和String
,哈希码功能不那么简单,因此不会保留顺序。
答案 1 :(得分:2)
HashMap没有排序,您的结果只是偶然的。试试这个
Map<Long, Long> m = new HashMap<>();
m.put(1001L, 1001L);
m.put(1000L, 1000L);
m.put(1002L, 1002L);
System.out.println(m);
输出
{1001=1001, 1000=1000, 1002=1002}
- 没有排序