这道题也是很经典的一道题了，408的算法题中也考过这个思想，因为两个链表已是升序，合并只需要两个指针，分别指向两个表的表头，分别比较两个指针所指向的结点的val，小的就插入到目标链表里面，再后移相应指针，直到有一个链表为空，再把剩下的非空链表全部查到目标链表后面即可

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution(object):
    def mergeTwoLists(self, list1, list2):
        """
        :type list1: Optional[ListNode]
        :type list2: Optional[ListNode]
        :rtype: Optional[ListNode]
        """
        ans = ListNode()
        current = ans
        i = list1
        j = list2
        while i is not None and j is not None:
            if i.val <= j.val:
                current.next = ListNode(i.val)
                i = i.next
            else:
                current.next = ListNode(j.val)
                j = j.next
            current = current.next
        current.next = i if i is not None else j
        return ans.next