將兩個(gè)升序鏈表合并為一個(gè)新的升序鏈表并返回。新鏈表是通過(guò)拼接給定的兩個(gè)鏈表的所有節(jié)點(diǎn)組成的犀斋。
示例：
輸入：1->2->4, 1->3->4
輸出：1->1->2->3->4->4
題解：一個(gè)為None后即可結(jié)束速缨，剩下的補(bǔ)充上
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
def mergeTwoLists(self, l1: ListNode, l2: ListNode) -> ListNode:
dummy = ListNode(-1)
head = dummy
while l1 and l2:
if l1.val <= l2.val:
head.next = ListNode(l1.val)
head = head.next
l1 = l1.next
else:
head.next = ListNode(l2.val)
head = head.next
l2 = l2.next
head.next = l1 if l1 else l2
return dummy.next
