Question

我编写了一个合并2个已排序链接列表的函数。但在我知道我提出的逻辑是否正确之前，我在调用函数时遇到了无限循环。我不确定为什么会这样。以下是代码：

ListNode *merge_sorted_list(ListNode *head1, ListNode *head2) {
    cout<<"hi\n";
    ListNode *curr = new ListNode(-1);
    ListNode *start = curr;
    cout<<"hi\n";
    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;

}

我无法打印＆＃39; hi＆＃39;在控制台窗口上。此外，head1和head2不是NULL值。请帮助。 更新：能够看到＆＃39; hi＆＃39;现在

Answer 1

首先，你正在破坏head1和head2因为你写了

curr->next = head1;

以后：

curr->next = head2;

所以你要混合两者，最后你可能会有一些记忆泄漏。

因此，您希望将两个列表组合到一个新列表中。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;
}

C ++：程序在运行函数时进入无限循环

1 个答案: