链表删除算法

Question

此程序用于从链表中删除节点。它工作正常，但问题是它打印0，而不是不打印已删除节点的值。例如，如果我的列表是1->2->3->-999。我想删除，比如说2。然后，最终列表打印为1-->0-->3-->-999。为什么是0？？？

enter code here
#include<stdio.h>
#include<malloc.h>
struct list {
        int number;
        struct list *next;
        };
typedef struct list node;
void create(node *);
void print(node *);
node *delete(node *,int);
main() {
int key;
node *head;
head = (node *)malloc(sizeof(node));
create(head);
printf("Your list as you entered is....................n");
print(head);
printf("Which element do you want to delete?n ");
scanf("%d",&key);
head = delete(head,key);
printf("The new list is ..............................n ");
print(head);
return 0;
}
void create(node *list) {
        printf("Enter a number,-999 to stop data entryingn");
        scanf("%d",&list->number);
        if(list->number == -999) {
                list->next = NULL;
        }
        else {
                list->next = (node *)malloc(sizeof(node));
                create(list->next);
        }
}
void print(node *list) {
        if(list->number != -999) {
                printf("%d-->",list->number);
                print(list->next); }
        else {
                printf("%d",list->number);
        }
}
node *delete(node *list,int key) {
        node *prev_ptr = NULL;
        node *curr_ptr;
        for(curr_ptr=list;curr_ptr!=NULL;prev_ptr=curr_ptr,curr_ptr=curr_ptr->next) {
        if(curr_ptr->number == key) {
                break;
        }
        }
        if(prev_ptr == NULL) {
                list = curr_ptr->next;
        }
        else {
                prev_ptr = curr_ptr->next;
                }
        free(curr_ptr);
        return(list);
}

Answer 1

在delete例程中重新排列指针时出错。代替

...
else {
    prev_ptr = curr_ptr->next;
}
...

你应该有

...    
else {
    prev_ptr->next = curr_ptr->next;
}
...

显示0是因为您释放了curr_ptr（因此它指向任何位置）。但真正的问题是你不应该把它列入清单。