我有一個(gè)簡單的問題：給定一個(gè)數(shù)字?jǐn)?shù)組，找到數(shù)組的 N 個(gè)最大數(shù)字，但我需要使用多線程來解決這個(gè)問題，比如使用 10 個(gè)線程。我不想對數(shù)組進(jìn)行排序：只需遍歷它，并將每個(gè)元素與大小為 N 的結(jié)果數(shù)組的最小值進(jìn)行比較（用 初始化Double.MIN_VALUE）。遍歷數(shù)組后，結(jié)果數(shù)組將包含我輸入數(shù)組的最大 N 個(gè)元素。對于多線程，我不希望每個(gè)線程都有一個(gè)結(jié)果數(shù)組，這樣我以后就不必合并它們。這就是為什么我希望所有線程都對共享結(jié)果數(shù)組進(jìn)行操作。我意識到這不是最好的解決方案，但我仍然想了解我應(yīng)該如何實(shí)現(xiàn)它。我試過這個(gè)，但它不起作用：public class Problem {private static final int MY_THREADS = 10;public static void main(String[] args) { double[] array = {...}; double[] maximums = new double[3]; for (int i = 0; i < maximums.length; ++i) { maximums[i] = Double.MIN_VALUE; } ExecutorService executor = Executors.newFixedThreadPool(MY_THREADS); Runnable worker = new MyRunnable(array, maximums); executor.execute(worker); executor.shutdown(); while (!executor.isTerminated()) { } System.out.println(Arrays.toString(maximums));}public static class MyRunnable implements Runnable { private double[] array; private double[] maximums; MyRunnable(double[] array, double[] maximums) { this.array = array; this.maximums = maximums; } @Override public void run() { int i = 0; while (i < array.length) { if (array[i] > maximums[getMinIndex(maximums)]) { maximums[getMinIndex(maximums)] = array[i]; } ++i; } }}private static int getMinIndex(double[] array) { int minIndex = -1; double min = Double.MAX_VALUE; for (int i = 0; i < array.length; ++i) { if (array[i] < min) { min = array[i]; minIndex = i; } } return minIndex;}}有人可以幫忙嗎？謝謝。
1 回答
泛舟湖上清波郎朗
TA貢獻(xiàn)1818條經(jīng)驗(yàn) 獲得超3個(gè)贊
我不知道你為什么要多線程這樣的東西，然后還要避免排序 - 但是suuuuuuureeeee。你可以這樣做：
class Problem {
private static final int MY_THREADS = 10;
private static final double[] maximums = new double[3];
public static void main(String[] args) {
double[] array = {...};
for ( int i = 0; i < maximums.length; ++i) {
maximums[i] = Double.MIN_VALUE; //Remember that this won't work with negative values in array
}
ExecutorService executor = Executors.newFixedThreadPool(MY_THREADS);
int start = 0;
int length = array.length/MY_THREADS;
for( int i = 0; i < MY_THREADS; i++ )
{
//You probably want to give it only part of array to consider,
//otherwise you are wasting resources and might even try to insert same element more than once.
Runnable worker = new MyRunnable(Arrays.copyOfRange( array, start, start + length ) );
executor.execute(worker);
start += length;
}
executor.shutdown();
while (!executor.isTerminated()) {
}
System.out.println( Arrays.toString( maximums ));
}
//This is unsynchronized - but with this problem - it will at worst cause unnecessary insert attempt.
private static int getMinIndex() {
int minIndex = -1;
double min = Double.MAX_VALUE;
for (int i = 0; i < maximums.length; ++i) {
if (maximums[i] < min) {
min = maximums[i];
minIndex = i;
}
}
return minIndex;
}
//You have to synchronize the insertion somehow,
// otherwise you might insert two values into same spot losing one of max.
private static synchronized void insertIntoMaximum( double k ){
int minIndex = getMinIndex();
if( maximums[minIndex] < k ){
maximums[minIndex] = k;
}
}
public static class MyRunnable implements Runnable {
private double[] array;
//Since its inner class, I didn't think passing maximums into it was necessary.
// You could pass it here, but you would still probably need to call parent for insertion.
MyRunnable(double[] array) {
this.array = array;
}
@Override
public void run() {
//this while was an interesting attempt at making indexed for, that doesn't even need to be indexed.
for( double v : array )
{
if( v > maximums[getMinIndex()] )
{
insertIntoMaximum( v );
}
}
}
}
}
我仍然可能會(huì)避免使用多線程，創(chuàng)建一個(gè)新線程可能非常昂貴 - 所以它很可能甚至不會(huì)節(jié)省時(shí)間，特別是考慮到您仍然需要同步插入。
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