将两个升序链表合并为一个新的 升序 链表并返回。新链表是通过拼接给定的两个链表的所有节点组成的。 

输入：l1 = [1,2,4], l2 = [1,3,4]
输出：[1,1,2,3,4,4]

示例 2：

输入：l1 = [], l2 = []
输出：[]

示例 3：

输入：l1 = [], l2 = [0]
输出：[0]

提示：

• 两个链表的节点数目范围是 [0, 50]
• -100 <= Node.val <= 100
• l1 和 l2 均按 非递减顺序 排列

  # Definition for singly-linked list.
  # class ListNode:
  #     def __init__(self, val=0, next=None):
  #         self.val = val
  #         self.next = next
  class Solution:
      def mergeTwoLists(self, list1: Optional[ListNode], list2: Optional[ListNode]) -> Optional[ListNode]:
          if(list1==None)://若有一个链表数值为空则返回
              return list2
          elif(list2==None):
              return list1
          elif(list1==None and list2==None):
              return None
          else://均不为空则排序
              list3 = ListNode(0,None)//list3为头结点
              list_head = list3//头指针
              list_temp1 = list1//temp1，2分别为list1，2的前驱节点
              list_temp2 = list2
              while (list_temp1!=None and list_temp2!=None):
                  if(list_temp1.val<=list_temp2.val):
                      list_head.next = ListNode(list_temp1.val,None)
                      list_head = list_head.next
                      list_temp1 = list_temp1.next
                  else:
                      list_head.next = ListNode(list_temp2.val,None)
                      list_head = list_head.next
                      list_temp2 = list_temp2.next
              while(list_temp1!=None):
                  list_head.next = ListNode(list_temp1.val,None)
                  list_head = list_head.next
                  list_temp1 = list_temp1.next
              while(list_temp2!=None):
                  list_head.next = ListNode(list_temp2.val,None)
                  list_head = list_head.next
                  list_temp2 = list_temp2.next    
              return list3.next