这段代码中存在一个问题，它删除了 cur->calendar 链表中的所有内容，而不仅仅是条件中的那个

Question

我尝试了所有方法，但我不知道现在该怎么做，所以非常感谢您的帮助！

void DeleteAtPosition(Appointment* head, int pos) {
    Appointment* cur = head->next;
    Appointment* prev = head;
    int i = 1;
    if (isEmptylist(head)) {
        cout << "LIST IS EMPTY" << endl;
    }
    if (pos == 0) {
        delete head;
        head = cur;
    }
    while (i < pos && cur->next != NULL) {
        cur = cur->next;
        prev = prev->next;
        i++;
    }
    prev->next = cur->next;
    delete cur;
}


void cancel(Company* c, string title) {
    
    Employee* cur = c->head;
    Appointment* curr=cur->Calendar;
    int pos = 0;
    while (cur != NULL) {
        while (cur->Calendar != NULL) {
            if (cur->Calendar->Title == title) {
                DeleteAtPosition(curr, pos);
            }
            pos++;
            cur->Calendar = cur->Calendar->next;
        }
    cur = cur->next;
    }
}

我认为问题出在取消功能上，而不是删除功能上，但我不知道如何解决

Answer 1

问题是你在做cur->Calendar = cur->Calendar->next;。您正在将原始指针移到列表上，因此在内部 while 循环的末尾，cur->Calendar 将变为 null.

您应该改为使用 curr 变量来遍历列表，如下所示：

void cancel(Company* c, string title) {
    Employee* cur = c->head;
    int pos = 0;
    while (cur != NULL) {
        Appointment* curr=cur->Calendar; // reset curr pointer in every iteration
        while (curr != NULL) { // use the curr pointer to go over the list
            if (curr->Title == title) { // check the title of the current appointment with the "curr" pointer
                DeleteAtPosition(cur->Calendar, pos); // use the original pointer as the head of the list for deletion
            }
            pos++;
            curr = curr->next; // move the "curr" pointer to the next element of the list
        }
        cur = cur->next;
    }
}

这段代码中存在一个问题，它删除了 cur->calendar 链表中的所有内容，而不仅仅是条件中的那个

there is a problem in this code its deleting everything in cur->calendar linked list not just the one in the condition

c++

singly-linked-list

doubly-linked-list