Question

我写了一个代码，获取带有数字的链表，并试图使该列表升序。不幸的是，代码没有被遵守，我也不知道为什么。

我尝试使用指针和引用，但是我不能把手放在哪里出了问题。

#include <iostream>
using namespace std;

class ListNode {
public:
  ListNode(const int &info) : data(info), nextPtr(0) {}

  int getData() const { return data; }
  ListNode *getNext() const { return nextPtr; }
  void setNext(ListNode *next) { nextPtr = next; }

private:
  int data;
  ListNode *nextPtr;
};

ListNode sort(ListNode &temp) {
  ListNode *first = &temp;
  ListNode *curr = first;
  ListNode *next = curr->getNext();
  ListNode *found = 0;

  while (curr->getNext() != 0) {
    if (curr->getData() > next->getData()) {
      if (curr == first) {
        first = next;
        found = curr;
      }

      else {
        curr->setNext(next->getNext());
        found = next;
      }

      break;
    }

    curr = next;
    next = next->getNext();
  }

  curr = first;
  next = curr->getNext();

  while (curr->getNext() != 0) {
    if (curr->getData() <= found->getData() &&
        found->getData() < next->getData()) {
      curr->setNext(found);
      found->setNext(next);
      break;
    }

    curr = next;
    next = next->getNext();
  }

  return *first;
}

void print(ListNode &temp) {
  ListNode *curr = &temp;

  while (curr != 0) {
    cout << curr->getData() << " ";
    curr = curr->getNext();
  }

  cout << endl;
}

int main1() {
  ListNode a(2);
  ListNode b(5);
  ListNode c(8);
  ListNode d(13);
  ListNode e(18);
  ListNode f(7);
  ListNode g(21);

  a.setNext(&b);
  b.setNext(&c);
  c.setNext(&d);
  d.setNext(&e);
  e.setNext(&f);
  f.setNext(&g);

  print(a);
  print(sort(a));

  return 0;
}

我检查了一百次，不知道为什么这段代码没有编译。

Answer 1

sort()应该返回指向该节点的指针，因此返回first而不是*first并将返回类型更改为ListNode*。然后将print(sort(a))更改为print(*sort(a))。看到它在这里运行：http://coliru.stacked-crooked.com/a/c3e72983e83f6914

Answer 2

     #include<iostream>
    using namespace std;

class ListNode
{
public:
    ListNode(const int &info) :data(info), nextPtr(0)
    {
    }

    int getData() const
    {
        return data;
    }
    ListNode * getNext() const
    {
        return nextPtr;
    }
    void setNext(ListNode * next)
    {
        nextPtr = next;
    }
private:
    int data;
    ListNode *nextPtr;
};

ListNode sort(ListNode &temp)
{
    ListNode *first = &temp;
    ListNode *curr = first;
    ListNode *next = curr->getNext();
    ListNode *found = 0;

    while (curr->getNext() != 0)
    {
        if (curr->getData() > next->getData())
        {
            if (curr == first)
            {
                first = next;
                found = curr;
            }

            else
            {
                curr->setNext(next->getNext());
                found = next;
            }

            break;
        }

        curr = next;
        next = next->getNext();
    }

    curr = first;
    next = curr->getNext();

    while (curr->getNext() != 0)
    {
        if (curr->getData() <= found->getData() && found->getData() < next->getData())
        {
            curr->setNext(found);
            found->setNext(next);
            break;
        }

        curr = next;
        next = next->getNext();
    }

    return *first;
}

您正在将临时（rvalue）传递给需要左值的函数

    void print(const ListNode &temp)
{


const ListNode * curr = &temp;

    while (curr != 0)
    {
        cout << curr->getData() << " ";
        curr = curr->getNext();
    }

    cout << endl;
}
//I am expecting main() here , I think this is what you meant.

int main()
{
    ListNode a(2);
    ListNode b(5);
    ListNode c(8);
    ListNode d(13);
    ListNode e(18);
    ListNode f(7);
    ListNode g(21);

    a.setNext(&b);
    b.setNext(&c);
    c.setNext(&d);
    d.setNext(&e);
    e.setNext(&f);
    f.setNext(&g);

    print(a);
    print(sort(a));

    return 0;
    }

代码无法编译-链表，列表按升序排列

2 个答案: