我在java中遇到了多线程合并排序算法中的一个问题。
我应该通过将原始数组划分为subArray来将代码修改为3,4,5,6,7,8线程合并排序。目前它有2 subArray s。
如何将原始数组拆分为3,4,5,6,7,8 subArray以实现我的目标？
此外，我应该编写更多方法，因为mergeSort方法目前调用lefthalf和righthalf方法。因此，对于3,4,5,6,7,8个线程，我应该编写其他方法。
我怎么处理这个？
two_threaded_merge_sort.java
public class two_threaded_merge_sort {
public static void finalMerge(int[] a, int[] b) {
int[] result = new int[a.length + b.length];
int i=0;
int j=0;
int r=0;
while (i < a.length && j < b.length) {
if (a[i] <= b[j]) {
result[r]=a[i];
i++;
r++;
} else {
result[r]=b[j];
j++;
r++;
}
if (i==a.length) {
while (j<b.length) {
result[r]=b[j];
r++;
j++;
}
}
if (j==b.length) {
while (i<a.length) {
result[r]=a[i];
r++;
i++;
}
}
}
}
public static void main(String[] args) throws InterruptedException {
Random rand = new Random();
int[] original = new int[9000000];
for (int i=0; i<original.length; i++) {
original[i] = rand.nextInt(1000);
}
long startTime = System.currentTimeMillis();
int[] subArr1 = new int[original.length/2];
int[] subArr2 = new int[original.length - original.length/2];
System.arraycopy(original, 0, subArr1, 0, original.length/2);
System.arraycopy(original, original.length/2, subArr2, 0, original.length - original.length/2);
Worker runner1 = new Worker(subArr1);
Worker runner2 = new Worker(subArr2);
runner1.start();
runner2.start();
runner1.join();
runner2.join();
finalMerge (runner1.getInternal(), runner2.getInternal());
long stopTime = System.currentTimeMillis();
long elapsedTime = stopTime - startTime;
System.out.println("2-thread MergeSort takes: " + (float)elapsedTime/1000 + " seconds");
}
}
Worker.java
class Worker extends Thread {
private int[] internal;
public int[] getInternal() {
return internal;
}
public void mergeSort(int[] array) {
if (array.length > 1) {
int[] left = leftHalf(array);
int[] right = rightHalf(array);
mergeSort(left);
mergeSort(right);
merge(array, left, right);
}
}
public int[] leftHalf(int[] array) {
int size1 = array.length / 2;
int[] left = new int[size1];
for (int i = 0; i < size1; i++) {
left[i] = array[i];
}
return left;
}
public int[] rightHalf(int[] array) {
int size1 = array.length / 2;
int size2 = array.length - size1;
int[] right = new int[size2];
for (int i = 0; i < size2; i++) {
right[i] = array[i + size1];
}
return right;
}
public void merge(int[] result, int[] left, int[] right) {
int i1 = 0;
int i2 = 0;
for (int i = 0; i < result.length; i++) {
if (i2 >= right.length || (i1 < left.length && left[i1] <= right[i2])) {
result[i] = left[i1];
i1++;
} else {
result[i] = right[i2];
i2++;
}
}
}
Worker(int[] arr) {
internal = arr;
}
public void run() {
mergeSort(internal);
}
}
非常感谢！
答案 0 :(得分：2)
需要有一个sort函数将数组分成k个部分，然后创建k个线程来对每个部分进行排序，使用自顶向下或自底向上的方法（自下而上会略快），并等待所有线程到完整。
此时有k个分类部分。这些可以使用k-way合并（复杂）一次合并，或者一次合并一对部分（2路合并），也许使用多个线程，但此时进程可能是内存带宽有限，所以多线程可能没什么用。
当将数组分成k个部分时，可以使用类似的东西来保持大小相似：
int r = n % k;
int s = n / k;
int t;
for each part{
t = r ? 1 : 0;
r -= t;
size = s + t;
}
或
int r = n % k;
int s = n / k + 1;
while(r--){
next part size = s; // n / k + 1
}
s -= 1;
while not done{
next part size = s; // n / k
}
答案 1 :(得分：1)
从我的角度来看，你的辛勤工作已经完成。现在你必须用线程数来参数化算法。
您的算法有两部分
还有两个组成部分：
关于主题
在我看来，Start / join方法在这种情况下没用，因为在完成所有线程之前，最后一次合并无法启动。我更喜欢“2路合并”（@rcgldr答案）和线程池（ExecutorService）。 您必须小心线程同步和共享内存。
总而言之，我提出了一个不同的解决方案：
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Iterator;
import java.util.List;
import java.util.Random;
import java.util.concurrent.Executors;
import java.util.concurrent.ExecutorService;
public class MultithreadedMergeSort {
private int[] array;
private int numThreads;
private List<int[]> sortedFragments;
private MultithreadedMergeSort(int numThreads, int[] array) {
this.numThreads = numThreads;
this.array = array;
}
// Basic algorithm: it sort recursively a fragment
private static void recursiveMergeSort(int[] array, int begin, int end) {
if (end - begin > 1) {
int middle = (begin + end) / 2;
recursiveMergeSort(array, begin, middle);
recursiveMergeSort(array, middle, end);
merge(array, begin, middle, end);
}
}
// Basic algorithm: it merges two consecutives sorted fragments
private static void merge(int[] array, int begin, int middle, int end) {
int[] firstPart = Arrays.copyOfRange(array, begin, middle);
int i = 0;
int j = middle;
int k = begin;
while (i < firstPart.length && j < end) {
if (firstPart[i] <= array[j]) {
array[k++] = firstPart[i++];
} else {
array[k++] = array[j++];
}
}
if (i < firstPart.length) {
System.arraycopy(firstPart, i, array, k, firstPart.length - i);
}
}
public static void sort(int[] array, int numThreads) throws InterruptedException {
if (array != null && array.length > 1) {
if (numThreads > 1) {
new MultithreadedMergeSort(numThreads, array).mergeSort();
} else {
recursiveMergeSort(array, 0, array.length);
}
}
}
private synchronized void mergeSort() throws InterruptedException {
// A thread pool
ExecutorService executors = Executors.newFixedThreadPool(numThreads);
this.sortedFragments = new ArrayList<>(numThreads - 1);
int begin = 0;
int end = 0;
// it split the work
for (int i = 1; i <= (numThreads - 1); i++) {
begin = end;
end = (array.length * i) / (numThreads - 1);
// sending the work to worker
executors.execute(new MergeSortWorker(begin, end));
}
// this is waiting until work is done
wait();
// shutdown the thread pool.
executors.shutdown();
}
private synchronized int[] notifyFragmentSorted(int begin, int end) {
if (begin > 0 || end < array.length) {
// the array is not completely sorted
Iterator<int[]> it = sortedFragments.iterator();
// searching a previous or next fragment
while (it.hasNext()) {
int[] f = it.next();
if (f[1] == begin || f[0] == end) {
// It found a previous/next fragment
it.remove();
return f;
}
}
sortedFragments.add(new int[]{begin, end});
} else {
// the array is sorted
notify();
}
return null;
}
private class MergeSortWorker implements Runnable {
int begin;
int end;
public MergeSortWorker(int begin, int end) {
this.begin = begin;
this.end = end;
}
@Override
public void run() {
// Sort a fragment
recursiveMergeSort(array, begin, end);
// notify the sorted fragment
int[] nearFragment = notifyFragmentSorted(begin, end);
while (nearFragment != null) {
// there's more work: merge two consecutives sorted fragments, (begin, end) and nearFragment
int middle;
if (nearFragment[0] < begin) {
middle = begin;
begin = nearFragment[0];
} else {
middle = nearFragment[0];
end = nearFragment[1];
}
merge(array, begin, middle, end);
nearFragment = notifyFragmentSorted(begin, end);
}
}
}
public static void main(String[] args) throws InterruptedException {
int numThreads = 5;
Random rand = new Random();
int[] original = new int[9000000];
for (int i = 0; i < original.length; i++) {
original[i] = rand.nextInt(1000);
}
long startTime = System.currentTimeMillis();
MultithreadedMergeSort.sort(original, numThreads);
long stopTime = System.currentTimeMillis();
long elapsedTime = stopTime - startTime;
// warning: Take care with microbenchmarks
System.out.println(numThreads + "-thread MergeSort takes: " + (float) elapsedTime / 1000 + " seconds");
}
}