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:
        # 如果其中一个链表为空，则直接返回另一个链表
        if not pHead1: return pHead2
        if not pHead2: return pHead1
        # 创建一个新的头结点，用于保存合并后的链表
        dummy = ListNode(0)
        current = dummy
        # 遍历两个链表，比较结点值大小，逐个将结点连接到合并链表中
        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
        # 将剩余的结点连接到合并链表的末尾
        if pHead1:
            current.next = pHead1
        if pHead2:
            current.next = pHead2
        # 返回合并后的链表的头结点
        return dummy.next