Question

我正在尝试实现一种算法来合并C ++中的两个链接列表，以便在hackerrank上解决相应的problem 。但是我收到了分段错误错误消息，我不知道为什么。这是我的代码：

Node* MergeLists(Node *headA, Node* headB)
{
  if (headA == NULL || headB == NULL) {
    return (headA == NULL ? headB : headA);
  }

  Node* currentA = headA; currentA->next = NULL;
  Node* nextA = headA->next; nextA->next = NULL;
  Node* currentB = headB; currentB->next = NULL;
  Node* nextB = currentB->next; nextB->next = NULL;

  while (currentA != NULL && currentB != NULL) {
    while (currentB != NULL && nextA != NULL && nextA->data <= currentB->data) {
      currentA = nextA;
      nextA = nextA->next;
    }
    if (currentA != NULL && currentA->data <= currentB->data) {
      currentA->next = currentB;
      currentA = nextA;
      if (nextA != NULL) {
        nextA = nextA->next;
      }
    }
    while (currentA != NULL && nextB != NULL && nextB->data <= currentA->data){
      currentB = nextB;
      nextB = nextB->next;
    }
    if (currentB != NULL && currentB->data <= currentA->data) {
      currentB->next = currentA;
      currentB = nextB;
      if (nextB != NULL) {
        nextB = nextB->next;
      }
    }
  }

  if (headA->data <= headB->data) {
    return headA;
  } else {
    return headB;
  }

}

错误消息如下：

GDB trace:
Reading symbols from solution...done.
[New LWP 2775]
Core was generated by `solution'.
Program terminated with signal SIGSEGV, Segmentation fault.
#0  0x0000000000400c94 in MergeLists (headA=headA@entry=0x1c46c30, 
    headB=headB@entry=0x1c46cb0) at solution.cc:32
32              ca -> next = cb;
#0  0x0000000000400c94 in MergeLists (headA=headA@entry=0x1c46c30, 
    headB=headB@entry=0x1c46cb0) at solution.cc:32
#1  0x0000000000400a0b in MergeLists (headB=0x1c46cb0, headA=0x1c46c30)
    at solution.cc:103
#2  main () at solution.cc:100

Answer 1

我认为你必须让你的指针“新”然后使用它们。当使用new运算符时，在堆上而不是在堆栈上声明对象。因此，您可以在函数中使用它们。我想你需要这样的东西：

  Node* currentA = new Node();currentA = headA; currentA->next = NULL;
  Node* nextA = new Node(); nextA = headA->next; nextA->next = NULL;
  Node* currentB = new Node(); currentB = headB; currentB->next = NULL;
  Node* nextB = new Node(); nextB = currentB->next; nextB->next = NULL;

我希望它会对你有所帮助。抱歉我的英文！

Answer 2

我猜分段错误是由解除引用的NULL指针引起的。

下一个指针的NULL初始化导致它。所有这些＆＃39; - ＆gt; next = NULL＆＃39;语句可以从下面的代码段中删除。它应该适用于合并2个链接列表，因为其余代码看起来没问题。

Node* currentA = headA; currentA->next = NULL;
Node* nextA = headA->next; nextA->next = NULL;
Node* currentB = headB; currentB->next = NULL;
Node* nextB = currentB->next; nextB->next = NULL;

希望它有所帮助！

合并两个链表算法分割故障C ++

2 个答案: