我有以下代码，其中我使用 HashMap（使用两个并行数组）来存储键值对（键可以有多个值）。现在，我必须存储和加载它以供将来使用，这就是我使用文件通道存储和加载它的原因。这段代码的问题是：我可以在我的 8 GB 服务器中存储近 1.2 亿个键值对（实际上，我可以为我的 JVM 分配 8 GB 中的近 5 GB，而这两个并行数组占用将近 2.5 GB，其他内存用于我的代码的各种处理）。但是，我必须存储近 600/7 亿个键值对。任何人都可以帮助我如何修改此代码，因此我可以存储近 600/7 亿个键值对。或者对此的任何评论对我来说都会很好。还有一点，我必须在内存中加载和存储哈希图。使用文件通道需要一点时间。根据 Stack Overflow 的各种建议，我没有找到更快的建议。我也使用过 ObjectOutputStream，压缩输出流，但是比下面的代码慢。无论如何以这种方式存储这两个并行数组，因此加载时间会快得多。我在下面的代码中给出了一个测试用例。对此的任何评论也将对我有所帮助。
import java.io.*;
import java.util.ArrayList;
import java.util.Iterator;
import java.util.Arrays;
import java.util.Random;
import java.nio.*;
import java.nio.channels.FileChannel;
import java.io.RandomAccessFile;
public class Test {
public static void main(String args[]) {
try {
Random randomGenerator = new Random();
LongIntParallelHashMultimap lph = new LongIntParallelHashMultimap(220000000, "xx.dat", "yy.dat");
for (int i = 0; i < 110000000; i++) {
lph.put(i, randomGenerator.nextInt(200000000));
}
lph.save();
LongIntParallelHashMultimap lphN = new LongIntParallelHashMultimap(220000000, "xx.dat", "yy.dat");
lphN.load();
int tt[] = lphN.get(1);
System.out.println(tt[0]);
} catch (Exception e) {
e.printStackTrace();
}
}
}
class LongIntParallelHashMultimap {
private static final long NULL = -1L;
private final long[] keys;
private final int[] values;
private int size;
private int savenum = 0;
private String str1 = "";
private String str2 = "";
public LongIntParallelHashMultimap(int capacity, String st1, String st2) {
keys = new long[capacity];
values = new int[capacity];
Arrays.fill(keys, NULL);
savenum = capacity;
str1 = st1;
str2 = st2;
}
public void put(long key, int value) {
int index = indexFor(key);
while (keys[index] != NULL) {
index = successor(index);
}
keys[index] = key;
values[index] = value;
++size;
}
public int[] get(long key) {
int index = indexFor(key);
int count = countHits(key, index);
int[] hits = new int[count];
int hitIndex = 0;
while (keys[index] != NULL) {
if (keys[index] == key) {
hits[hitIndex] = values[index];
++hitIndex;
}
index = successor(index);
}
return hits;
}
private int countHits(long key, int index) {
int numHits = 0;
while (keys[index] != NULL) {
if (keys[index] == key) {
++numHits;
}
index = successor(index);
}
return numHits;
}
private int indexFor(long key) {
return Math.abs((int) ((key * 5700357409661598721L) % keys.length));
}
private int successor(int index) {
return (index + 1) % keys.length;
}
public int size() {
return size;
}
public void load() {
try {
FileChannel channel2 = new RandomAccessFile(str1, "r").getChannel();
MappedByteBuffer mbb2 = channel2.map(FileChannel.MapMode.READ_ONLY, 0, channel2.size());
mbb2.order(ByteOrder.nativeOrder());
assert mbb2.remaining() == savenum * 8;
for (int i = 0; i < savenum; i++) {
long l = mbb2.getLong();
keys[i] = l;
}
channel2.close();
FileChannel channel3 = new RandomAccessFile(str2, "r").getChannel();
MappedByteBuffer mbb3 = channel3.map(FileChannel.MapMode.READ_ONLY, 0, channel3.size());
mbb3.order(ByteOrder.nativeOrder());
assert mbb3.remaining() == savenum * 4;
for (int i = 0; i < savenum; i++) {
int l1 = mbb3.getInt();
values[i] = l1;
}
channel3.close();
} catch (Exception e) {
System.out.println(e);
}
}
public void save() {
try {
FileChannel channel = new RandomAccessFile(str1, "rw").getChannel();
MappedByteBuffer mbb = channel.map(FileChannel.MapMode.READ_WRITE, 0, savenum * 8);
mbb.order(ByteOrder.nativeOrder());
for (int i = 0; i < savenum; i++) {
mbb.putLong(keys[i]);
}
channel.close();
FileChannel channel1 = new RandomAccessFile(str2, "rw").getChannel();
MappedByteBuffer mbb1 = channel1.map(FileChannel.MapMode.READ_WRITE, 0, savenum * 4);
mbb1.order(ByteOrder.nativeOrder());
for (int i = 0; i < savenum; i++) {
mbb1.putInt(values[i]);
}
channel1.close();
} catch (Exception e) {
System.out.println("IOException : " + e);
}
}
}