题目如下:
题目
这道题比较简单，需要注意的是如果输入有空链表，那么直接返回非空链表。如果都为空链表那么返回空。思路为将两个链表都装入到一个list中，通过sorted进行排序，然后根据排序后的list返回新的链表，参考代码如下：
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
def mergeTwoLists(self, l1, l2):
"""
:type l1: ListNode
:type l2: ListNode
:rtype: ListNode
"""
if l1 is None and l2 is None:
return None
elif l1 is not None and l2 is None:
return l1
elif l2 is not None and l1 is None:
return l2
l = []
while l1 is not None:
l.append(l1.val)
l1 = l1.next
while l2 is not None:
l.append(l2.val)
l2 = l2.next
l = sorted(l)
new_l = ListNode(l[0])
head_l = new_l
for i in range(1,len(l)):
new_l.next = ListNode(l[i])
new_l = new_l.next
new_l.next = None
return head_l
ps：如果您有好的建议，欢迎交流 :-D，也欢迎访问我的个人博客：tundrazone.com
