链接列表删除算法

Question

该程序用于从链接列表中删除节点。它工作正常，但问题是它打印0而不是打印删除节点的值。 例如。如果我的列表是1 - > 2 - > 3 - > - 999.并且我想删除说2.然后最终列表打印为1 - > 0 - > - > 3 - > - 999.为什么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 entrying\n"); 
     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

您在重新排列删除例程中的指针时出错。取而代之的

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

你应该有

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

的0被显示，因为您免费curr_ptr（因此它指向无处）。但真正的问题是你不应该把它列入清单。