要先定义1个虚拟头结点以免判空
class Solution {
public:
ListNode* merge(ListNode* l1, ListNode* l2) {
auto dummy = new ListNode(-1), tail = dummy; // 定义一个虚拟头结点
while (l1 && l2)
{
if (l1->val < l2->val)
{
tail = tail->next = l1; // 答案链表的尾结点
l1 = l1->next;
}
else
{
tail = tail->next = l2;
l2 = l2->next;
}
}
// 2个链表有可能会有1个非空
if (l1) tail->next = l1;
if (l2) tail->next = l2;
return dummy->next; // 返回虚拟头结点的next
}
};