从C中的Singly Linked List中删除前端节点

Question

我正试图从Singly Linked List中删除，但当我试图从第一个元素中删除时，它会打印垃圾。我认为问题来自delete_node函数，然而，我尝试了所有的方法，但都无法解决。

    #include <stdio.h>//prinf
#include <stdlib.h>//alloc mallco callo
typedef struct node node;
 struct node{
    int number;
    node *next;
};
node *new_node(int num){
    node *n= (node*) malloc(sizeof(node));
    n->number=num;
    n->next=NULL;
    return n;
}
void node_free_all(node *n){
    if(n != NULL){
        node_free_all(n->next);
        free(n);
    }
}
void print_nodes(node *n){
    if(n != NULL){
        print_nodes(n->next);
        printf("Number is: %dn",n->number);
    }
}
void delete_node(node *n, int num){
    node *rmNode= (node*)malloc(sizeof(node));
    //delete first
    if( n!= NULL && n->number==num){
        rmNode = n;
        n=n->next;
        free(rmNode);
    }
    //all but first
    while(n != NULL){
        if(n->next != NULL && n->next->number == num){
            rmNode= n->next;
            n->next = rmNode->next;
            free(rmNode);
            break;
        }
        n=n->next;
    }   
}
int main(){
    int i;
    node *head= (node*) malloc(sizeof(node));
    node *curr;
    head=NULL;
   for(i=1;i<=10;i++) {
      curr = new_node(i);
      curr->next =head;
      head=curr;
    }

    printf("Everything:n");
    print_nodes(head);
    printf("Deleting 1:n");
    delete_node(head,1);
    print_nodes(head);
    printf("Deleting 5:n");
    delete_node(head,5);
    print_nodes(head);
    printf("Deleting 2:n");
    delete_node(head,2);
    print_nodes(head);
    printf("Deleting 3:n");
    delete_node(head,3);
    print_nodes(head);
    printf("Deleting 10:n");
    delete_node(head,10);
    print_nodes(head);
    printf("Deleting 9:n");
    delete_node(head,9);
    print_nodes(head);
node_free_all(head);
// node_free_all(list);
return 0;
}

我做错了什么？

Answer 1

您不需要为rmNode分配内存。另外，您需要将head指针的引用传递给函数delete_node，因为每次更新列表时，如果您必须删除列表的第一个元素，那么在这种情况下，head指针也会被更新。

struct node
{
    int number;
    node *next; 
};
void delete_node(struct node** head_ref, int num)
{
    struct node* temp;
    struct node* current = (*head_ref);

    //delete first
    if( current != NULL && current->number == num)
    {
        temp = current;
        current = current->next;
        free(temp);
        (*head_ref) = current;
    }
    else
    {
        //all but first
        while(current != NULL)
        {
            if(current->next != NULL && current->next->number == num)
            {
                temp = current->next;
                current->next = temp->next;
                free(temp);
                break;
            }
            current = current->next;
        }
    }
}

Answer 2

这里有一个版本，它删除了所有与num匹配的项，并始终更新头项。

void delete_node(node** head_ref, int num)
{
    node* temp;
    node* last = 0;
    node* current = *head_ref;
    while(current)
    {       
        if( current->number == num )
        {
            temp = current;
            if( current == *head_ref )
                current = (*head_ref) = current->next;
            else
                current = last->next = current->next;
            free(temp);
        } else {
            last = current;
            current = current->next;
        }
    }
}

你必须称之为delete_node(&head,1);
因为它需要一个标题项的地址才能将其更改为