从链表问题 C 中删除节点

Question

>我只是想从链表中删除一个节点，似乎遇到了麻烦。我想知道是否有人可以看看可能出了什么问题？谢谢！

struct ets {
    struct node *equip_head;
    struct node *member_head;
    struct node *loan_head;
    const char *equip_fname;
    const char *member_fname;
    const char *loan_fname;
};
struct node {
    void *data; /* Accepts all data, yay */
    struct node *next;
};
BOOLEAN deleteMember(struct ets *ets, char MemberID[]) {
    struct node *current = ets->member_head;
    struct node *tmpNode = current;
    struct member_data *member_data = NULL;
    while (current != NULL) {
        member_data = current->data;
        if (strcmp(member_data->MemberID, MemberID) == 0) {
            tmpNode = current;
            current = current->next;
            free(tmpNode->data);
            free(tmpNode);
            return TRUE;
        }
        current = current->next;
    }
    return FALSE;
}

Answer 1

您不会从列表中删除节点。您可以执行此操作以从列表中删除节点：

BOOLEAN deleteMember(struct ets *ets, char MemberID[]) {
    struct node *current = ets->member_head;
    struct node *prev=NULL;
    struct member_data *member_data = NULL;

    while(current != NULL) {
        member_data = current->data;
        if(strcmp(member_data->MemberID, MemberID) == 0) {
            if(prev==NULL) // removing 1st node
                ets->member_head=current->next;
            else
                prev->next=current->next; // removing current node from list
            free(current->data);
            free(current);
            return TRUE;
        }
        prev = current;
        current = current->next;
    }
    return FALSE;
}

Answer 2

因为你有一个链表，你删除算法是坏的。这是您目前正在做的事情：

• 找到要删除的成员。好。你有node-1 -> node -> node+1
• 删除成员。为什么不呢。但是您的列表将变得node-1 -> unallocated node并且无法找到下一个节点

相反，您应该测试下一个节点是否具有正确的 id，以具有类似以下内容的内容：prev_node(current) -> node_to_delete -> next_node'

然后你可以做：

tmpNode = current->next;
current->next = tmpNode->next; /* ok for the chaining */
free(tmpNode->data); /* deletion will be ok */
free(tmpNode);

当然，对第一个和最后一个节点进行了特殊管理...

编辑：阿里已经给出了答案。我留下这个作为对OP算法为什么被破坏的评论