Question

我无法用一组固定长度整数的基数排序来绕头。在我下面尝试实现最低有效数字基数排序时，我有一个名为num_at的函数，它返回数字num中的数字d。我写的代码有w = 2，其中w代表每个数字的长度。（基本上，这个代码是为3位数字编写的，如下面的输入所示）。

我对每个数字的键索引计数进行了建模，但是我得到了[0, 12, 32, 32, 44, 0]的输出，并且坦率地说，为什么会遇到困难。这是我的输入：[32,12,99,44,77, 12]我用99个值启动了count数组，因为根据键索引计数，所有可能的值都可以在0到99之间。这里有什么东西立刻跳出来不正确吗？另外，我的num_at方法是正确的方法吗？还是有更好的方法？

def num_at(num, d)
    return num if num/10 == 0 
    a = num 
    #return the dth digit of num
    counter = 0 
    until counter == d 
        a/=10 
        counter += 1 
    end 

    a % 10
end

def radix_sort(arr)
    w = 2
    #count_arr can have any possible number from 0-99
    aux = Array.new(arr.length) {0}

    d = w-1
    while d >= 0 
        count = Array.new(99) {0}
        i = 0 

        #create freq arr
        while i < arr.length 
            count[num_at(arr[i], d) + 1] += 1 #offset by 1 
            i += 1 
        end 

        #compute cumulates 
        i = 0 
        while i < count.length - 1
            count[i + 1] += count[i]
            i += 1
        end 

        z = 0 
        #populate aux arr
        while z < arr.length 
            aux[num_at(arr[z], d)] = arr[z]
            count[num_at(arr[z], d)] += 1 
            z += 1
        end

        #override original arr
        z = 0 
        while z < arr.length 
            arr[z] = aux[z]
            z += 1
        end 
        d -= 1
    end

    arr
end

这是从我试图用Ruby实现的教科书中获取的LSD基数排序的Java代码：

public static void lsd(String[] a)
{
 int N = a.length;
 int W = a[0].length;
 for (int d = W-1; d >= 0; d--)
 {
 int[] count = new int[R];
 for (int i = 0; i < N; i++)
 count[a[i].charAt(d) + 1]++;
 for (int k = 1; k < 256; k++)
 count[k] += count[k-1];
 for (int i = 0; i < N; i++)
 temp[count[a[i].charAt(d)]++] = a[i];
 for (int i = 0; i < N; i++)
 a[i] = temp[i];
 }

和密钥索引计数的伪代码（对每个字符重复）：

Task: sort an array a[] of N integers between 0 and R-1
Plan: produce sorted result in array temp[]
 1. Count frequencies of each letter using key as index
 2. Compute frequency cumulates
 3. Access cumulates using key as index to find record positions.
 4. Copy back into original array
 5. List item

LSD radix sort. Consider characters d from right to left  Stably sort
using dth character as the key via key-indexed counting.

上述java代码/伪代码信息来自此链接：http://www.cs.princeton.edu/courses/archive/spring07/cos226/lectures/11RadixSort.pdf

LSD Radix排序整数

0 个答案: