# 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