在你的实现中，我并没有看出特别低效的地方。我承认我不是很理解你的哈希/查找策略，但如果你说在你的情况下它表现良好，我会相信你的。
唯一可能有所改善的是将键从Entry的值数组中移出。
不要这样做：
class Entry {
long[] values;
}
if ( entry.values[key_index] == key ) {
试试这个：
class Entry {
long key;
long values[];
}
if ( entry.key == key ) {
与其付出访问数组成员的代价，再进行边界检查，最后获取数组中的值，你应该只付出访问成员的代价。
是否有一种比数组更快的随机访问数据类型？
我对这个问题很感兴趣，所以我建立了一个测试环境。这是我的数组接口:
interface Array {
long get(int i);
void set(int i, long v);
}
这个“Array”在索引超出边界时会有未定义的行为。我做了一个显而易见的实现：
class NormalArray implements Array {
private long[] data;
public NormalArray(int size) {
data = new long[size];
}
@Override
public long get(int i) {
return data[i];
}
@Override
public void set(int i, long v) {
data[i] = v;
}
}
然后是控制器：
class NoOpArray implements Array {
@Override
public long get(int i) {
return 0;
}
@Override
public void set(int i, long v) {
}
}
最后，我设计了一个“数组”，其中前10个索引是硬编码成员。通过开关设置/选择成员：
class TenArray implements Array {
private long v0;
private long v1;
private long v2;
private long v3;
private long v4;
private long v5;
private long v6;
private long v7;
private long v8;
private long v9;
private long[] extras;
public TenArray(int size) {
if (size > 10) {
extras = new long[size - 10];
}
}
@Override
public long get(final int i) {
switch (i) {
case 0:
return v0;
case 1:
return v1;
case 2:
return v2;
case 3:
return v3;
case 4:
return v4;
case 5:
return v5;
case 6:
return v6;
case 7:
return v7;
case 8:
return v8;
case 9:
return v9;
default:
return extras[i - 10];
}
}
@Override
public void set(final int i, final long v) {
switch (i) {
case 0:
v0 = v; break;
case 1:
v1 = v; break;
case 2:
v2 = v; break;
case 3:
v3 = v; break;
case 4:
v4 = v; break;
case 5:
v5 = v; break;
case 6:
v6 = v; break;
case 7:
v7 = v; break;
case 8:
v8 = v; break;
case 9:
v9 = v; break;
default:
extras[i - 10] = v;
}
}
}
我用这个测试工具进行了测试:
import java.util.Random;
public class ArrayOptimization {
public static void main(String[] args) {
int size = 10;
long[] data = new long[size];
Random r = new Random();
for ( int i = 0; i < data.length; i++ ) {
data[i] = r.nextLong();
}
Array[] a = new Array[] {
new NoOpArray(),
new NormalArray(size),
new TenArray(size)
};
for (;;) {
for ( int i = 0; i < a.length; i++ ) {
testSet(a[i], data, 10000000);
testGet(a[i], data, 10000000);
}
}
}
private static void testGet(Array a, long[] data, int iterations) {
long nanos = System.nanoTime();
for ( int i = 0; i < iterations; i++ ) {
for ( int j = 0; j < data.length; j++ ) {
data[j] = a.get(j);
}
}
long stop = System.nanoTime();
System.out.printf("%s/get took %fms%n", a.getClass().getName(),
(stop - nanos) / 1000000.0);
}
private static void testSet(Array a, long[] data, int iterations) {
long nanos = System.nanoTime();
for ( int i = 0; i < iterations; i++ ) {
for ( int j = 0; j < data.length; j++ ) {
a.set(j, data[j]);
}
}
long stop = System.nanoTime();
System.out.printf("%s/set took %fms%n", a.getClass().getName(),
(stop - nanos) / 1000000.0);
}
}
结果有些令人惊讶。对于小于等于10的尺寸，TenArray的性能比NormalArray更快。减去开销（使用NoOpArray平均值），你会发现TenArray只需普通数组时间的约65%。所以，如果你知道数组的最大可能大小，我想你可以超越数组的速度。我想switch语句使用的边界检查要么更少，要么更高效，而不是数组。
NoOpArray/set took 953.272654ms
NoOpArray/get took 891.514622ms
NormalArray/set took 1235.694953ms
NormalArray/get took 1148.091061ms
TenArray/set took 1149.833109ms
TenArray/get took 1054.040459ms
NoOpArray/set took 948.458667ms
NoOpArray/get took 888.618223ms
NormalArray/set took 1232.554749ms
NormalArray/get took 1120.333771ms
TenArray/set took 1153.505578ms
TenArray/get took 1056.665337ms
NoOpArray/set took 955.812843ms
NoOpArray/get took 893.398847ms
NormalArray/set took 1237.358472ms
NormalArray/get took 1125.100537ms
TenArray/set took 1150.901231ms
TenArray/get took 1057.867936ms
现在，我不确定你是否可以在实践中获得比数组更快的速度；显然，这种方式会产生与接口/类/方法相关的任何开销。
