c - 冒泡排序双向链表

Question

我的双向链表冒泡排序函数有问题。当我以单链接方式(仅使用 ->next)对节点进行排序时它可以工作，但我无法使用 ->prev 指针。这是我正在使用的代码:

void sort(int count)
{
    struct data *tmp,*current,*nextone;
    int i,j;
    for(i=0;i<count;i++)
    {
        current = first;
        for(j=0;j<count-1-i;j++ )
        {
            if(current->number > current->next->number)
            {
                nextone = current->next;
                current->next = nextone->next;
                nextone->next = current;
                if(current == first)
                {
                    first = nextone;
                    current = nextone;
                }
                else
                {
                    current = nextone;
                    tmp->next = nextone;
                }
            }
            tmp = current;
            current = current->next;
        }
    }
}

这是我正在使用的结构(使用列表的第一个和最后一个元素的全局变量):

struct data    
{
    int id;
    char name[20];
    int number;
    struct data *next;
    struct data *prev;
};

struct data *first = NULL;
struct data *last = NULL;

Answer 1

以下逻辑可行。

我会遵循类似的算法......如果你想移动整个节点......

struct data *before, *after;
if(current->number > current->next->number)
{
    before = current->prev;
    after = current->next;
    if(before != NULL){
        before->next = after;
    }
    current->next = after->next;
    current->prev = after;
    after->next = current;
    after->previous = before;
}

或者，如果目的是对数据进行排序，您可以简单地交换节点中的数字，而不必费心移动整个节点。您可以扩展以下逻辑以包括 char 数组和 id 的交换。

if(current->number > current->next->number)
{
    int tempNum = current->number;
    current->number = current->next->number;
    current->next->number = tempNum;
}