您的实施中没有什么让我觉得效率特别低。我承认我并没有真正遵循您的散列/查找策略，但是如果您说它在您的情况下表现出色，我会相信您。
我唯一期望可能会有所不同的是将键移出Entry.
而不是这样：
class Entry {
long[] values;
}
//...
if ( entry.values[key_index] == key ) { //...
试试这个：
class Entry {
long key;
long values[];
}
//...
if ( entry.key == key ) { //...
而不是产生访问成员的成本，加上进行边界检查，然后获取数组的值，您应该只产生访问成员的成本。
是否有比数组更快的随机访问数据类型？
我对这个问题的答案很感兴趣，所以我搭建了一个测试环境。这是我的数组接口：
interface Array {
long get(int i);
void set(int i, long v);
}
当索引超出范围时，此“数组”具有未定义的行为。我把明显的实现放在一起：
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;
}
}
然后是一个控件：
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 个索引是硬编码的成员。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;
}
}
}
我用这个线束测试了它：
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);
}
}
结果有些出人意料。TenArray 的执行速度比 NormalArray 快得多（对于大小 <= 10）。减去开销（使用 NoOpArray 平均值），您会得到 TenArray 的时间约为普通数组的 65%。因此，如果您知道数组的可能最大大小，我想可能会超过数组的速度。我想 switch 使用比数组更少的边界检查或更有效的边界检查。
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，我不确定您是否可以在实践中获得比阵列更快的速度；显然，这种方式会产生与接口/类/方法相关的任何开销。