Question

因此，对于输入大小-10、20、30，以下算法可以很好地发挥作用
输入大小-输出中可见35个时滞
输入大小-40您将等待一段时间
大于40的内容-无法完成排序。

public class TestSorting {

    public static void main(String[] args) {

        long start, end;

        Comparable[] numbers = generateRandom(45);
        Comparable[] temp;

        temp = Arrays.copyOf(numbers,numbers.length);
        System.out.println("-----------------------------------------------------------");
        System.out.println("                  M E R G E   S O R T");
        System.out.println("-----------------------------------------------------------");
        System.out.println("Initial: " + Arrays.asList(temp));
        start = System.nanoTime();
        Merge.sort(temp);
        end = System.nanoTime();
        System.out.println("Final  : " + Arrays.asList(temp));
        System.out.println("Total Runtime: " + (end-start));


    }

    private static Comparable[] generateRandom(int size) {
        Comparable[] data = new Comparable[size];

        Random random = new Random();

        for(int i = 0 ;i<size;i++)
            data[i] = random.nextInt(100);

        return data;
    }
}

Merge.java

public class Merge {

    public static boolean lessThan(Comparable x, Comparable y) {
       return x.compareTo(y)<0;
    }

    public static void sort(Comparable[] data) {
        sort(data,0,data.length-1);
    }


    private static void sort(Comparable[] data,int low,int high) {
        if(high<=low) return;

        int mid = low + (high-low)/2;
        sort(data,0,mid);
        sort(data,mid+1,high);
        merge(data,low,mid,high);
    }

    private static void merge(Comparable[] data, int low, int mid, int high) {

        Comparable[] X = new Comparable[mid-low+1]
        Comparable[] Y = new Comparable[high-mid];

        for(int i=0;i<X.length;i++) X[i] = data[low+i];
        for(int i=0;i<Y.length;i++) Y[i] = data[mid+1+i];

        int start1 = 0, start2 = 0;
        while(start1<X.length && start2<Y.length) {
            if(lessThan(X[start1],Y[start2])) data[low++] = X[start1++];
            else data[low++] = Y[start2++];
        }

        while (start1 < X.length) data[low++] = X[start1++];
        while (start2 < Y.length) data[low++] = Y[start2++];
    }

}

任何人都可以帮助我找出问题所在吗？是因为额外的本地数组创建使它运行缓慢？

Answer 1

递归调用

sort(data,0,mid);

实际上应该是

sort(data,low,mid);

这段运行了很长时间的合并排序代码有什么问题？

1 个答案: