c++ - 代码不编译 - 链表，按升序排序列表

Question

我写了一个代码来获取一个带有数字的链表，并试图使该列表成为升序序列。不幸的是代码不符合，我不知道为什么。

我尝试过使用指针和引用，但我无法确定哪里出了问题。

#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;
}

我已经检查了数百次，但不知道为什么这段代码无法编译。

Answer 1

sort() 应该返回指向节点的指针，因此返回 first 而不是 *first 并将返回类型更改为 列表节点*。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