将两个有序链表合并为一个新的有序链表并返回。新链表是通过拼接给定的两个链表的所有节点组成的。 示例：输入：1->2->4, 1->3->4 输出：1->1->2->3->4->4
方法一：递归
递归地定义在两个链表里的 merge 操作：
list1[0]+merge(list1[1:],list2) list1[0]<list2[0]
list2[0]+merge(list1,list2[1:])? otherwise?
递归过程：判断 l1 和 l2 哪一个的头元素更小，然后递归地决定下一个添加到结果里的值。如果两个链表都是空的，那么过程终止，所以递归过程最终一定会终止。
class Solution:
def mergeTwoLists(self, l1, l2):
if l1 is None:
return l2
elif l2 is None:
return l1
elif l1.val < l2.val:
l1.next = self.mergeTwoLists(l1.next, l2)
return l1
else:
l2.next = self.mergeTwoLists(l1, l2.next)
return l2
方法二：迭代
class Solution:
def mergeTwoLists(self, l1, l2):
# maintain an unchanging reference to node ahead of the return node.
prehead = ListNode(-1)
prev = prehead
while l1 and l2:
if l1.val <= l2.val:
prev.next = l1
l1 = l1.next
else:
prev.next = l2
l2 = l2.next
prev = prev.next
# exactly one of l1 and l2 can be non-null at this point, so connect
# the non-null list to the end of the merged list.
prev.next = l1 if l1 is not None else l2
return prehead.next
原文：https://www.cnblogs.com/yinminbo/p/11880676.html