ComparisonCountingSort伪代码难度

Question

这个算法在课堂上给我们：

当输入是数组A[60,35,81,98,14,47]时，我在尝试确定输出时遇到了问题。

我想习惯伪代码语法，所以我尝试将其转换为Java语法，但生成的程序不会显示结果。这是我转换的代码：

public class Driver
{
    public static void main(String[] args)
    {
        int[] test = {60, 35, 81, 98, 14, 47}; 
        int[] converse = new int[6];

        converse = countSort(test);

        for(int i : converse)
        {
            System.out.println(converse[i]);
        }
    }

    public static int[] countSort(int[] a)
    {
        int[] s = new int[6];
        int[] count = new int[6];
        int n = a.length + 1;

        for(int i = 0; i < (n-1);)
        {
            count[i] = 0;
        }

        for(int i = 0; i < (n-2);)
        {
            for(int j = (i+1); i < (n-1);)
            {
                if(a[i] < a[j])
                {
                    count[j] = count[j]+1;//What
                }
                else
                {
                    count[i] = count[i]+1;
                }
            }
        }

        for (int i = 0; i < (n-1);)
        {
            s[count[i]] = a[i];
        }

        return s;
    }
}

Answer 1

int[] test = {60,35,81,98,14,47};从未使用过。

您的意思是将test传递给countSort吗？并且也对它的结果做了一些事情。

虽然我没有真正看过算法，但这里还有其他问题：

int n = a.length + 1; for (int i = 0; i < (n - 1);) { count[i] = 0; }

这将无限循环。 i永远不会有任何影响。

Answer 2

1. Your array test is never used.
2. In your for loops, i never gets incremented. It will loop infinetely.
3. Instead of n = a.length+1; just use n = a.length. it makes your code easier to read. You then also have to check all your loops.

When you are running the code with a testset larger or less than 6 integer values your code wont run correctly or even fail with an ArrayOutOfBounds Exception, as you allocate the arrays s and count for only 6 values. Try int[] count = new int[a.length]; instead of int[] count = new int[a.length];. The same for s.