# 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:
# write code here
# 整体思路:
# 创建一个哑节点作为最终链表的前置节点
# 比较链表1和链表2的头节点,将较小的那一个接入最终链表中
# 更新最终链表的尾节点和链表1和链表2的头节点
# 如果有一个链表的节点已经全部加入最终链表中
# 将剩下链表的剩下节点接入最终链表
# 返回哑节点的下一个节点
dummy = ListNode(0)
cur = dummy
while pHead1 and pHead2:
if pHead1.val >= pHead2.val:
cur.next = pHead2
pHead2 = pHead2.next
else:
cur.next = pHead1
pHead1 = pHead1.next
cur = cur.next
if pHead1:
cur.next = pHead1
else:
cur.next = pHead2
return dummy.next