c++:程序在运行函数时进入无限循环

Question

我写了一个函数来合并2个排序的链表。但在我知道我提出的逻辑是否正确之前，我遇到了一个无限循环，只要函数被调用。我不知道为什么会这样。下面是代码:

ListNode *merge_sorted_list(ListNode *head1, ListNode *head2) {
    cout<<"hin";
    ListNode *curr = new ListNode(-1);
    ListNode *start = curr;
    cout<<"hin";
    while(head1 || head2) {
        if(head1->val == head2->val) {
            curr->next = head1;
            curr = curr->next;
            curr->next = head2;
            if(head1) head1 = head1->next;
            if(head2) head2 = head2->next;
        } else if(head1->val < head2->val) {
            curr->next = head1;
            curr = curr->next;
            if(head1) head1 = head1->next;
        } else {
            curr->next = head2;
            curr = curr->next;
            if(head2) head2 = head2->next;
        }
    }
    return start->next;
}

我无法在控制台窗口上打印"hi"。另外，head1和head2不是NULL值。请帮助。更新:现在能够看到"嗨"。

Answer 1

首先，你破坏了head1和head2，因为你写了

curr->next = head1;

,后来:

curr->next = head2;

所以你把两者混合在一起，最后你可能会有一些内存泄漏。

所以你想把两个列表合并成一个新列表。所以试试这个:

ListNode *merge_sorted_list(ListNode *head1, ListNode *head2) 
{
    ListNode* pReturn = new ListNode();
    ListNode* tempRet = pReturn;
    ListNode* temp1 = head1;
    ListNode* temp2 = head2;
    while(temp1 != nullptr || temp2 != nullptr)
    {
        if(temp1 && temp2)
        {
            if(temp1->val < temp2->val)
            {
                tempRet->next = new ListNode(temp1); //the constructor should copy the ListNode value
                tempRet = tempRet->next;
                temp1 = temp1->next;
            }
            else if(temp1->val > temp2->val)
            {
                tempRet->next = new ListNode(temp2);
                tempRet = tempRet->next;
                temp2 = temp2->next;
            }
            else //doubled value
            {
                temp1 = temp1->next;
            }
        }
        else if(temp1)
        {
            while(temp1)
            {
                tempRet->next = new ListNode(temp1);
                tempRet = tempRet -> next;
                temp1 = temp1->next;
            }
        }
        else if(temp2)
        {
            while(temp2)
            {
                tempRet->next = new ListNode(temp2);
                tempRet = tempRet -> next;
                temp2 = temp2->next;
            }
        }
    }
    return pReturn;
}