题解 | #合并两个排序的链表#

# 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
        # 创建一个哨兵节点，便于处理头节点的合并
        dummy = ListNode(0)
        current = dummy

        # 当两个链表均不为空时，比较当前节点值并连接较小的节点
        while pHead1 and pHead2:
            if pHead1.val <= pHead2.val:
                current.next = pHead1
                pHead1 = pHead1.next
            else:
                current.next = pHead2
                pHead2 = pHead2.next
            current = current.next
        
        # 连接剩余的节点（如果有的话）
        if pHead1:
            current.next = pHead1
        elif pHead2:
            current.next = pHead2

        # 返回合并后的链表（跳过哨兵节点）
        return dummy.next