C++ 排序链表的合并-a

已知两个已排序链表头节点指针l1与l2，将这两个链表合并，合并后认为有序的，返回合并后的头节点。
思路：
比较l1和l2指向的节点，将较小的节点插入到pre指针后，并向前移动较小节点对应的指针。

#include <stdio.h>
struct ListNode
{
 int val;
 ListNode* next;
 ListNode(int x) : val(x), next(NULL) {}
};
class Solution
{
public:
 Solution() {}
 ~Solution() {}
 ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
  ListNode temp_head(0);
  ListNode* pre = &temp_head;
  while (l1 && l2)
  {
   if (l1->val < l2->val)
   {
    pre->next = l1;
    l1 = l1->next;
   }
   else
   {
    pre->next = l2;
    l2 = l2->next;
   }
   pre = pre->next;
  }
  if (l1)
  {
   pre->next = l1;
  }
  if (l2)
  {
   pre->next = l2;
  }
  return temp_head.next;
 }
};
int main()
{
 ListNode a(1);
 ListNode b(4);
 ListNode c(6);
 ListNode d(0);
 ListNode e(5);
 ListNode f(7);
 a.next = &b;
 b.next = &c;
 d.next = &e;
 e.next = &f;
 Solution solve;
 ListNode* head = solve.mergeTwoLists(&a, &d);
 while (head)
 {
  printf("%d\n", head->val);
  head = head->next;
 }
 return 0;
}

运行结果：
0
1
4
5
6
7

来源：CSDN

作者：Gianna K

链接：https://blog.csdn.net/weixin_44208324/article/details/104524647