Question

所以我试图在C中对一个双向链表进行冒泡排序。我希望交换整个节点而不仅仅是每个节点内的元素。但是，当内循环即将停止时，我会遇到分段错误。我甚至在纸上画了我的代码，但我仍然无法弄清楚它有什么问题。下面是我的排序功能。

typedef struct list{
    char letter;
    double probability;
    struct list* previous;
    struct list* next;
}List;

void sort(int n){
    int i, j;
    struct list *nxt, *current;
    if (current==NULL){
        return;
    }
    for(i=0;i<n;i++){
        current = head;
        for(j=0;j<n-1-i;j++){
            nxt = current->next;
            if (current->probability<nxt->probability){ 
                if(current->previous!=NULL){
                    current->previous->next=nxt;
                }
                if(nxt->next!=NULL){
                    nxt->next->previous=current;
                }
                current->next=nxt->next;
                nxt->next=current;
                nxt->previous=current->previous;
                current->previous= nxt;
                nxt->previous->next=nxt;
                current->next->previous=current;
                printf("swapped\n");
            }
            current = current->next;
        }
    }
}

Answer 1

尝试更简单地编写代码：

@WebMvcTest

我认为您的错误与行... if (current->probability<nxt->probability){ struct list *prev = current->previous; struct list *next_next = nxt->next; swap(current, next); current->previous = prev; current->next = next; next->previous = current; next->next = next_next; if (prev != NULL) { prev->next = current; } if (next_next != NULL) { next_next->previous = next; } printf("swapped\n"); } current = current->next; ...有关，因为如果您交换元素，则应使用current = current->next。

函数交换可能如下所示：

current->next->next

C中的冒泡排序双重链表

1 个答案: