Radix分类功能出现问题

Question

In my program I am using a Radix Sort via Linked List to sort nine digit numbers, but for some reason it is not sorting correctly.

This is how I generate my numbers:

void genData(int *dta, int n) 
{
    // generate the numbers at random
    for(int i=0; i < n; i++)
        dta[i] =  rand()%889 + 111 + 1000*(rand()%889 + 111) + 1000000*(rand()%889 + 111);
}

This is the Radix Sort function: The outer loop is run through 3 times. Once for each set of 3 digits.

int radixSort(int *dta, int n, int *out)
{ 
    // the dta array contains the data to be sorted.
    // n is the number of data items in the array
    // out is the array to put the sorted data

    node *bucket[1000]; 
    int count = 0; 
    for(int i = 0; i < n; i++)out[i] = dta[i]; 

    for (int pass = 0; pass < 3; pass++)  // outer loop
    {
        for(int j = 0; j < 1000; j++) // set bucket[] to all zeroes (NULL) for each pass 
        {
            bucket[j] = NULL;
        }

        for(int i = 0; i < n; i++) // inner loop -- walks through the out array (which contains the data to be sorted)
        {
            int index = 0; 
            int tmp = 0;
            switch(pass) 
            {
                case 0:
                    index = out[i] % 1000;
                    break;
                case 1:
                    tmp = out[i]/1000; // tmp = 123456
                    index = tmp%1000; // mid = 456// set index to the middle 3 digits
                    break;
                case 2:
                    tmp = out[i]/1000;  // set index to the first 3 digits
                    index = tmp/1000;
                    break;
            };

            //Create new head node if nothing is stored in location
            if(bucket[index] == NULL)           
            {   
                node *newNode = new node(0, bucket[0]); 
            }
            else
            {
                node *newNode =  new node(out[i], NULL); //Created new node, stores out[i] in it
                node *temp = bucket[index];
                while(temp->next != NULL) // finds the tail of the Linked List
                {
                    temp = temp->next;
                    count++; //For Big-O
                }
                temp->next = newNode;   // make tail point to the new node.
            }
            count++; //For Big-O
        } // end of the inner (i) loop

        int idx = 0; // for loading the out array
        for(int i = 0; i < 1000; i++)  // walk through the bucket
        {
            if(bucket[i] == NULL)continue; // nothing was stored here so skip to the next item

            // something is stored here, so it is put into the out array starting at the beginning (idx)
            out[idx++] = bucket[i]->data;

            if(bucket[i]->next->next != NULL || bucket[i]->next->next)
            // now see if there are more nodes in the linked list that starts at bucket[i]. If there are, put their data into out[idx++]
            {
                out[idx++] = bucket[i]->data;
            }
            count++; //For Big-O
        }


    }// end of the outer loop pass). The output (out) from this pass becomes the input for the next pass

    return count; // Again -- for Big-O 
}

I think that the problem might have to be with the new node that I create. What am I doing wrong?

Answer 1

Your logic for storing a number in a linked list is not correct.

Here is a suggested outline:

• Always create a new node to store the number.
• Always set the next pointer of the new node to NULL .

• Find the end of the linked list at bucket[index] .

  • If there is no linked list at bucket[index] then you have already found the end.

     node *newNode = new node(out[i], NULL); if (bucket[index] == NULL) { // there was no linked list there before; start one now. bucket[index] = newNode; } else { // find tail of linked list and append newNode node *temp = bucket[index]; while (temp->next != NULL) { temp = temp->next; count++; //For Big-O } temp->next = newNode; // make tail point to the new node. } 

EDIT: You already had a while loop that follows the linked list from its head to its tail.

To get the values out of the linked list, you also start at the head and then follow the list until you reach the tail. But, as you visit each node in the list, you get out a value.

            if (bucket[i] == NULL)continue; // nothing was stored here so skip to the next item

            // if we reach this point there is at least one value stored here

            // get values out
            node *temp = bucket[i];
            out[idx++] = temp->data;
            while (temp->next != NULL)
            {
                temp = temp->next;
                out[idx++] = temp->data;
            }

But we can make this cleaner with a do / while loop. You use a do / while when you want to do something at least once, and possibly more than once. In this case, if we run this loop at all it is because we want to get out at least one number. So:

            if (bucket[i] == NULL)continue; // nothing was stored here so skip to the next item

            // if we reach this point there is at least one value stored here

            // get values out
            node *temp = bucket[i];
            do
            {
                out[idx++] = temp->data;
                temp = temp->next;
            }
            while (temp != NULL);

It's cleaner to use a do / while loop, than to repeat the line that stores a value in out .

The loop can handle any length list other than length 0, and you already have the check with the continue to handle the case where there is no linked list in bucket[i] .