单链表冒泡排序

今天做链表排序有个误区，就是以为交换的时候要连next节点也交换，还要固定head节点，想了很久也没做出来，但是后来看网上的提示，才知道只要交换节点内的数据就可以了，根本不用交换next节点

#include <stdio.h>
#include <stdlib.h>
 
struct node
{
    int data;
    struct node *next;
};
 
struct node *create_list(int a[],int len)
{
    struct node *phead;
    struct node *ptr;
    struct node *pre;
    phead=(struct node *)malloc(sizeof(struct node));
    int i=0;
    phead->data=a[i];
    phead->next=NULL;
    ptr=phead->next;
    pre=phead;
    for(i=1;i<len;i++)
    {
        ptr=(struct node *)malloc(sizeof(struct node));
        ptr->data=a[i];
        ptr->next=NULL;
        pre->next=ptr;
        ptr=ptr->next;
        pre=pre->next;
    }
     
    return phead;
}
 
void print_list(struct node *phead)
{
    struct node *ptr=phead;
     
    while(ptr != NULL)
    {
        printf("%d ",ptr->data);
        ptr=ptr->next;
    }
     
    printf("\n");
}
 
struct node *bubble(struct node *phead,int len)
{
    struct node *ptr,*next;
    int temp;
     
    for(int i=0;i<len;i++)
    {
        ptr=phead;
        next=ptr->next;
        for(int j=len-i-1;j>0;j--)
        {
            if(ptr->data > next->data)
            {
                temp=ptr->data;
                ptr->data=next->data;
                next->data=temp;
            }
            ptr=ptr->next;
            next=next->next;
        }
    }
     
    return phead;
}
 
int main()
{
    int a[10]={
        5,3,6,8,9,6,5,4,2,7
    };
     
    struct node *phead;
    phead=create_list(a,10);    
 
    print_list(phead);
     
    phead=bubble(phead,10);
     
    print_list(phead);
}