使用C ++进行基数排序

Question

Suppose I have bunch of numbers. I have to first put the least significant digit into the corresponding bucket. Ex: 530 , I have to first put into the bucket 0. For number 61, I have to put into bucket 1.

I planned to use a multidimensional array to do this. So I create a 2-dimenional array, which nrows is 10 ( for 0~ 9) and ncolumns is 999999 ( because I don't know how large will the list be):

    int nrows = 10;
    int ncolumns = 999999;

    int **array_for_bucket = (int **)malloc(nrows * sizeof(int *));
         for(i = 0; i < nrows; i++)
             array_for_bucket[i] = (int *)malloc(ncolumns * sizeof(int));

    left = (a->value)%10;
    array_for_bucket[left][?? ] = a->value;

Then I created one node call a. In this node a, there is a value 50. To find out which bucket I want to put it in, I calculate "left" and I got 0. So I want to put this a-> value into bucket 0. But now I am stuck. How do I put this value into the bucket? I have to use a pointer array to do this.

I thought for a long time but still couldn't find a good way to do it. So please share some ideas with me. thank you!

Answer 1

There is a much easier way of doing this, and instead of radix*nkeys space you only need an nkeys -sized buffer.

Allocate a second buffer that can fit nkeys keys. Now do a first pass through your data and simply count how many keys end up in each bucket. You now can create a radix -sized array of pointers where each pointer is to the start of that bucket in the output buffer. Finally, the second pass though the data moves the keys. Every time you move a key, increment that bucket pointer.

Here's some C code to make into C++:

void radix_sort(int *keys, int nkeys)
{
    int *shadow = malloc(nkeys * sizeof(*keys));

    int bucket_count[10];
    int *bucket_ptrs[10];
    int i;

    for (i = 0; i < 10; i++)
        bucket_count[i] = 0;

    for (i = 0; i < nkeys; i++)
        bucket_count[keys[i] % 10]++;

    bucket_ptrs[0] = shadow;
    for (i = 1; i < 10; i++)
        bucket_ptrs[i] = bucket_ptrs[i-1] + bucket_count[i-1];

    for (i = 0; i < nkeys; i++)
        *(bucket_ptrs[keys[i] % 10]++) = keys[i];

    //shadow now has the sorted keys

    free(shadow);
}

But I may have misunderstood the question. If you are doing something a little different than radix sort, pleas add some details.

Answer 2

如果要存储指针，请查看Boost Pointer容器库。

Answer 3

C++ isn't my forte but this code from wikipedia-Raidx Sort is very comprehensive and probably is more C++-ish than what you've implemented so far. Hope it helps

Answer 4

This is C++, we don't use malloc anymore. We use containers. A two-dimensional array is a vector of vectors.

vector<vector<int> > bucket(10);
left = (a->value)%10;
bucket[left].push_back(a->value);