1

我无法按数字顺序将结构插入到链表中。每个结构都有一个“数字”，如下面的代码所示。我试图让编号最小的结构成为列表的头部（即：成为“人”指向的结构）。我整天都在盯着这段代码，我不知道它有什么问题。任何帮助深表感谢。谢谢

Person *addPerson(Person *people, Person *addition, int &numList)
{
if (people == NULL && numList == 0)
{
    people = addition;
    numList++;
    return people;
}
if (people->number >= addition->number)
{
    addition->nextPerson = people;
    people = addition;
    return people;
}
else if (people->number < addition->number && people->nextPerson != NULL)
{
    addPerson(people->nextPerson, addition, numList);
}
else if (people->number < addition->number && people->nextPerson == NULL)
{
    people->nextPerson = addition;
    numList++;
    return people;
}
}

编辑**

int main()
{
Person *blake = new Person;
Person *kyra = new Person;
Person *elon = new Person;
Person *bill = new Person;
Person *people = NULL;

blake->number = 1;
blake->name = "blake";
blake->lastName = "madden";
blake->nextPerson = NULL;

kyra->number = 2;
kyra->name = "kyra";
kyra->lastName = "madden";
kyra->nextPerson = NULL;

elon->number = 3;
elon->name = "elon";
elon->lastName = "musk";
elon->nextPerson = NULL;

bill->number = 4;
bill->name = "bill";
bill->lastName = "gates";
bill->nextPerson = NULL;

int num = 0;
int &numList = num;

people = addPerson(people, blake, numList);

people = addPerson(people, kyra, numList);
people = addPerson(people, elon, numList);
people = addPerson(people, bill, numList);

cout << people->name << '\n' << people->lastName;
}
4

3 回答 3

2

您没有使用addPerson()第三个if块中的返回值。尝试：

people->nextPerson = addPerson(people->nextPerson, addition, numList);
return people;

您还需要return people;在其中，否则您将在函数结束时运行并且不会返回任何明智的内容（我的编译器警告我，您应该更改警告设置，以便您的也这样做）。

通过上述更改，您的代码似乎可以正常运行。

于 2012-07-14T01:33:09.163 回答
1

您永远不会分配返回的头部。它应该是

people->nextPerson = addPerson(people->nextPerson, addition, numList);
于 2012-07-14T01:35:26.400 回答
0

你的问题是函数的参数。您希望以递归调用顺序修改前一个调用者的节点中的指针，但您只是传递下一个节点的指针值。这不允许您实际修改前一个节点的指针成员。

更改您的功能，使其看起来像这样：

void addPerson(Person** people, Person *addition, int &numList)

然后当你调用addPerson第三个块时，这样做：

addPerson(&(*(people)->nextPerson), addition, numList);

最后，当您访问people参数时，请确保执行额外的取消引用，例如：

*people = addition;

或者

if ((*people)->number >= addition->number)
于 2012-07-14T01:33:21.273 回答