我正在尝试使用合并排序实现多线程。我让它在将数组切成两半的位置制作新线程 该数组的排序取决于: [数组的大小] vs [我创建新线程的次数] 例如:如果我让它在大小为70的数组上只创建两个线程,那么数组将被排序,但如果我让它创建6,它将返回未排序。我认为可能有一件事是线程没有同步,但我使用了threadName.join()
这里有一些代码:merge.java
import java.util.Random;
public class merge implements Runnable {
int[] list;
int length;
int countdown;
public merge(int size, int[] newList, int numberOfThreadReps, int firstMerge) {
length = size;
countdown = numberOfThreadReps;
list = newList;
if (firstMerge == 1)
threadMerge(0, length - 1);
}
public void run() {
threadMerge(0, length - 1);
}
public void printList(int[] list, int size) {
for (int i = 0; i < size; i++) {
System.out.println(list[i]);
}
}
public void regMerge(int low, int high) {
if (low < high) {
int middle = (low + high) / 2;
regMerge(low, middle);
regMerge(middle + 1, high);
mergeJoin(low, middle, high);
}
}
public void mergeJoin(int low, int middle, int high) {
int[] helper = new int[length];
for (int i = low; i <= high; i++) {
helper[i] = list[i];
}
int i = low;
int j = middle + 1;
int k = low;
while (i <= middle && j <= high) {
if (helper[i] <= helper[j]) {
list[k] = helper[i];
i++;
} else {
list[k] = helper[j];
j++;
}
k++;
}
while (i <= middle) {
list[k] = helper[i];
k++;
i++;
}
helper = null;
}
public void threadMerge(int low, int high) {
if (countdown > 0) {
if (low < high) {
countdown--;
int middle = (low + high) / 2;
int[] first = new int[length / 2];
int[] last = new int[length / 2 + ((length % 2 == 1) ? 1 : 0)];
for (int i = 0; i < length / 2; i++)
first[i] = list[i];
for (int i = 0; i < length / 2 + ((length % 2 == 1) ? 1 : 0); i++)
last[i] = list[i + length / 2];
merge thread1 = new merge(length / 2, first, countdown, 0);// 0
// is
// so
// that
// it
// doesn't
// call
// threadMerge
// twice
merge thread2 = new merge(length / 2
+ ((length % 2 == 1) ? 1 : 0), last, countdown, 0);
Thread merge1 = new Thread(thread1);
Thread merge2 = new Thread(thread2);
merge1.start();
merge2.start();
try {
merge1.join();
merge2.join();
} catch (InterruptedException ex) {
System.out.println("ERROR");
}
for (int i = 0; i < length / 2; i++)
list[i] = thread1.list[i];
for (int i = 0; i < length / 2 + ((length % 2 == 1) ? 1 : 0); i++)
list[i + length / 2] = thread2.list[i];
mergeJoin(low, middle, high);
} else {
System.out.println("elsd)");
}
} else {
regMerge(low, high);
}
}
}
proj4.java
import java.util.Random;
public class proj4 {
public static void main(String[] args) {
int size = 70000;
int threadRepeat = 6;
int[] list = new int[size];
list = fillList(list, size);
list = perm(list, size);
merge mergy = new merge(size, list, threadRepeat, 1);
// mergy.printList(mergy.list,mergy.length);
for (int i = 0; i < mergy.length; i++) {
if (mergy.list[i] != i) {
System.out.println("error)");
}
}
}
public static int[] fillList(int[] list, int size) {
for (int i = 0; i < size; i++)
list[i] = i;
return list;
}
public static int[] perm(int[] list, int size) {
Random generator = new Random();
int rand = generator.nextInt(size);
int temp;
for (int i = 0; i < size; i++) {
rand = generator.nextInt(size);
temp = list[i];
list[i] = list[rand];
list[rand] = temp;
}
return list;
}
}
所以TL; DR我的数组没有根据数组的大小和我使用线程拆分数组的次数进行多线程合并排序...为什么会这样?
答案 0 :(得分:4)
哇。这是一部关于受虐狂的有趣练习。我相信你已经离开了,但我想为后代......
代码中的错误在mergeJoin
中,带有middle
参数。这适用于regMerge
但在threadMerge
传入的中间位置是(low + high) / 2
而不是(length / 2) - 1
。由于threadMerge
low
始终为0且high
为length - 1
且first
数组的大小为(length / 2)
。这意味着对于具有奇数条目的列表,它通常会失败,具体取决于随机化。
还有一些样式问题使得这个程序变得更加复杂和容易出错:
list.length
调用时,代码会传递一个大小的数组,这样会更直接,更安全。length/2
)。Merge
而不是merge
)firstMerge
应该是布尔值Thread
变量merge1
和merge
变量thread1
。杯。merge
的{{1}}构造函数很奇怪。我会在threadMerge(0,length -1)
的{{1}}回拨后拨打该电话。然后可以移除new
。proj4
超过最大值而不是最大值。我们倾向于认为firstMerge
超过high
。然后,代码可以for (int i = 0; i < 10; i++)
从i <= 9
转到j
,low
从< middle
转到k
。更好的对称性。祝你好运。