无法从单向链表中删除

Question

所以我有这个结构:

struct Book {
    char *title;
    char *author;
    int  pages;
    struct Book *next;
}*start=NULL;

我有以下用于删除的功能:

void delete(int x) {
    struct Book *current;
    for(current = start; current; current = current->next) {
        if(current->pages == x) {
            current = current->next;
            break;
        }
    }
}

出于某种原因，这不起作用。我做错了什么？

Answer 1

首先了解从链表中删除节点的逻辑。 假设你有一个链接列表 x1->x2->x3->x4 并且你想从中删除 x3。所以要删除x3，你所要做的就是让x2的next指针指向x4。但在你这样做之前，你应该释放分配给节点 x3 的内存。

下面是实现这个的简单代码:

void delete(int x) {
struct Book *current;
for(current = start; current; current = current->next) {
if(current->next != NULL){
if(current->next->pages == x) {
    Struct Book * temp = current->next; //save the address of the node to be deleted
    current->next = current->next->next; //remove the node from the linked list
    free(temp->author);
    free(...); //whatever memory you want to release
    free(temp);   //free the memory
    break;
   }   
 }
}
}

上面的实现不包括要删除的节点是链表的头节点或根节点的情况。

void delete(int x) {
if(start == NULL)
    return; 
struct Book *current;
if(start->pages == x)
{
    Struct Book * temp = start; //save the address of the node to be deleted
    start = start->next; //remove the node from the linked list
    free(temp->author);
    free(...); //whatever memory you want to release
    free(temp); 
    return;
}
for(current = start; current; current = current->next)
{
if(current->next != NULL){
if(current->next->pages == x) {
Struct Book * temp = current->next; //save the address of the node to be deleted
current->next = current->next->next; //remove the node from the linked list
free(temp->author);
free(...); //whatever memory you want to release
free(temp);   //free the memory
break;
}   
}

}