c - 从C中的链表中删除最后两个元素

Question

我有这段代码，它从链表中删除最后一个元素。我必须进行哪些更改才能删除链表的最后两个元素？

void deletesEnd() {
    struct node *temp, *last;

    temp = head;
    last = temp;

    while (temp != NULL && temp->next != NULL) {
        last = temp;
        temp = temp->next;
    }

    if (last == temp) {
        free(temp);
        head = NULL;
    } else {
        free(last->next);
        last->next = NULL;
    }
}

Answer 1

删除列表最后 2 个元素的最简单解决方案是调用 deletesEnd() 两次。请注意，deletesEnd() 应将 head 作为参数并返回新值。您将通过发出嵌套调用来删除最后 2 个:

struct node *deletesEnd(struct node *head) {
    struct node *temp, *last;

    last = temp = head;
    while (temp && temp->next != NULL) {
        last = temp;
        temp = temp->next;
    }
    if (last == head) {
        free(head);
        head = NULL;
    } else {
        free(last->next);
        last->next = NULL;
    }
    return head;
}

删除最后一个元素:head = deletesEnd(head);

删除最后 2 个元素:head = deletesEnd(deletesEnd(head));

设计的简单性足以补偿两次枚举列表的开销。

如果你绝对想要一个特定的功能，你可以这样扩展你的方法:

struct node *deleteLast2Nodes(struct node *head) {
    struct node *temp, *last;

    last = temp = head;
    while (temp && temp->next != NULL && temp->next->next != NULL) {
        last = temp;
        temp = temp->next;
    }
    if (last == head) {
        if (head) {
            free(head->next);
        }
        free(head);
        head = NULL;
    } else {
        free(last->next->next);
        free(last->next);
        last->next = NULL;
    }
    return head;
}