链表中的出队

发布于 2025-01-04 05:44:10 字数 758 浏览 2 评论 0原文

我正在尝试让我的出队方法在 LinkedList ADT 的实现上发挥作用。但是，它是从队列的开头而不是末尾删除。有什么帮助吗？我是 C 新手，正在尝试将 java 练习移植到 C.. 它应该删除列表的最后一个节点。

这是我的出队方法：

static void linkedQueueDequeue(Queue* q) {
    LinkedQueue *lq = ((LinkedQueue*)q->privateData);
    Node* temp = lq->head->next;
    lq->head->data = lq->head->next->data;
    lq->head->next = temp->next;
    free(temp);
    lq->size--;


}

这是尝试使最后一个节点出队时的输出：

=====================
|Testing LinkedQueue|
=====================
adding 1 to first node
adding 2 to second node
adding 3 to third node
adding 4 to fourth node
[1,2,3,4]
dequeue last node
should print [1,2,3]
[2,3,4]
return first node
peek: 2
what's the size?
size: 3

原文

I'm trying to get my dequeue method working on my implementation of a LinkedList ADT. However, it is removing from the beginning of the queue instead of the end. Any help with this? I'm new to C, and am trying to port a java exercise over to C.. It's supposed to remove the last node of the list.

Here's my dequeue method:

static void linkedQueueDequeue(Queue* q) {
    LinkedQueue *lq = ((LinkedQueue*)q->privateData);
    Node* temp = lq->head->next;
    lq->head->data = lq->head->next->data;
    lq->head->next = temp->next;
    free(temp);
    lq->size--;


}

Here's the output when trying to dequeue last node:

=====================
|Testing LinkedQueue|
=====================
adding 1 to first node
adding 2 to second node
adding 3 to third node
adding 4 to fourth node
[1,2,3,4]
dequeue last node
should print [1,2,3]
[2,3,4]
return first node
peek: 2
what's the size?
size: 3

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旧时光的容颜 2025-01-11 05:44:10

正如您已经看到的，linkedQueueDequeue 中的代码会弹出第一个条目，就像您想要一个堆栈 (LIFO) 一样，您可以将您的 temp 迭代到列表的末尾，然后删除它的 temp- >next：

static void linkedQueueDequeue(Queue* q) {
    LinkedQueue *lq = ((LinkedQueue*)q->privateData);
    Node* temp = lq->head->next;
    while(temp->next) temp = temp->next;
    free(temp->next);
    temp->next = 0;
    lq->size--;
}

另请注意，考虑到第 2 行中的转换 (LinkedQueue*)q，您的 Queue/LinkedQueue 实现有些奇怪。您确定需要该转换吗？我真的无法判断，因为您没有给我们 Queue 和 LinkedQueue 的定义。 LinkedQueue 中是否还有一个 ->tail ？如果是这样，那么您不需要迭代，可以使用 ->tail 来定位 temp （当然：您必须更新 ->tail 到新的结尾）。

As you saw already, the code in linkedQueueDequeue pops the first entry as if you wanted a stack (LIFO), you can iterate your temp to the end of the list, then remove it's temp->next:

static void linkedQueueDequeue(Queue* q) {
    LinkedQueue *lq = ((LinkedQueue*)q->privateData);
    Node* temp = lq->head->next;
    while(temp->next) temp = temp->next;
    free(temp->next);
    temp->next = 0;
    lq->size--;
}

Also note, that there ist something slightly odd about your Queue/LinkedQueue implementation considering the conversion (LinkedQueue*)q in line 2. Are you sure you need that cast? I cannot really tell because you did not give us the definitions of Queue and LinkedQueue. Is there maybe also a ->tail in LinkedQueue? If so, then you dont need the iteration and can instead use ->tail to position temp (and of course: you have to update ->tail to the new end).

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