题目要求:

思路:

思路1：递归
思路2：额外列表排序
思路3：常规遍历

代码如下:

思路1代码：

# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None
#
# 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
#
# 
# @param pHead1 ListNode类 
# @param pHead2 ListNode类 
# @return ListNode类
#
class Solution:def Merge(self , pHead1: ListNode, pHead2: ListNode) -> ListNode:if not pHead1: # pHead1为None时将pHead2整体返回return pHead2if not pHead2: # pHead2为None时将pHead1整体返回return pHead1if pHead1.val <= pHead2.val: # 值小的在前，后面指向第二小的值即可pHead1.next = self.Merge(pHead1.next,pHead2)return pHead1 # 将比较得到的小值返回else:pHead2.next = self.Merge(pHead1,pHead2.next)return pHead2# write code here

思路2代码：

# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None
#
# 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
#
# 
# @param pHead1 ListNode类 
# @param pHead2 ListNode类 
# @return ListNode类
#
class Solution:def Merge(self , pHead1: ListNode, pHead2: ListNode) -> ListNode:if not pHead1:return pHead2if not pHead2:return pHead1node_list = []while pHead1:node_list.append(pHead1)pHead1 = pHead1.nextwhile pHead2:node_list.append(pHead2)pHead2 = pHead2.nextnode_list.sort(key=lambda x:x.val)pHead = index = node_list[0]for item in node_list[1:]:index.next = itemindex = index.nextreturn pHead# write code here

思路3代码：

# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None
#
# 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
#
# 
# @param pHead1 ListNode类 
# @param pHead2 ListNode类 
# @return ListNode类
#
class Solution:def Merge(self , pHead1: ListNode, pHead2: ListNode) -> ListNode:if not pHead1:return pHead2if not pHead2:return pHead1pHead = index = ListNode(-1)while pHead1 and pHead2:if pHead1.val <= pHead2.val:index.next = pHead1pHead1 = pHead1.nextelse:index.next = pHead2pHead2 = pHead2.nextindex = index.nextif pHead1:index.next = pHead1else:index.next = pHead2return pHead.next# write code here