原题
https://leetcode.com/problems/merge-two-sorted-lists/description/
题干
Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists.
Example:
Input: 1->2->4, 1->3->4 Output: 1->1->2->3->4->4
这是凑字数的这是凑字数的这是凑字数的这是凑字数的这是凑字数的这是凑字数的这是凑字数的这是凑字数的这是凑字数的这是凑字数的这是凑字数的这是凑字数的这是凑字数的这是凑字数的这是凑字数的这是凑字数的这是凑字数的这是凑字数的这是凑字数的这是凑字数的这是凑字数的这是凑字数的这是凑字数的这是凑字数的这是凑字数的这是凑字数的这是凑字数的这是凑字数的这是凑字数的这是凑字数的这是凑字数的这是凑字数的这是凑字数的这是凑字数的这是凑字数的这是凑字数的这是凑字数的这是凑字数的这是凑字数的
解(官方)
代码语言：javascript复制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解(官方)
代码语言：javascript复制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解
代码语言：javascript复制class Solution(object):
def mergeTwoLists(self, l1, l2):
"""
:type l1: ListNode
:type l2: ListNode
:rtype: ListNode
"""
if not l1:
return l2
if not l2:
return l1
dummy = cur = ListNode(-1)
while l1 and l2:
if l1.val < l2.val:
cur.next = l1
l1 = l1.next
else:
cur.next = l2
l2 = l2.next
cur = cur.next
cur.next = l1 if l1 else l2
return dummy.next 
