Question

在我的程序中，我使用基于链接列表的基数排序来排序九位数字，但由于某种原因，它没有正确排序。

这是我生成数字的方式：

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);
}

这是Radix排序功能：外循环运行3次。每组3个数字一次。

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 
}

我认为问题可能与我创建的新节点有关。我做错了什么？

Answer 1

将数字存储在链表中的逻辑不正确。

以下是建议大纲：

始终创建一个新节点来存储号码。
始终将新节点的next指针设置为NULL。

在bucket[index]找到链接列表的结尾。

如果bucket[index]没有链接列表，那么您已经找到了结果。

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.
}

编辑：你已经有一个while循环，它从头到尾跟在链表之后。

要从链接列表中获取值，您也可以从头部开始，然后按照列表进行操作，直至到达尾部。但是，当您访问列表中的每个节点时，您会得到一个值。

            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;
            }

但我们可以通过do / while循环使其更清洁。当您想要至少执行一次，可能不止一次时，可以使用do / while。在这种情况下，如果我们完全运行这个循环，那是因为我们想要输出至少一个数字。所以：

            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);

使用do / while循环比使用out重复存储值的行更清晰。

循环可以处理除长度0以外的任何长度列表，并且您已经使用continue进行检查以处理bucket[i]中没有链接列表的情况。

Radix排序功能有问题

1 个答案: