题解 | #合并两个排序的链表#递归解法

# 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:
        # 使用递归
        if not pHead1:
            return pHead2
        if not pHead2:
            return pHead1
        # 原链表作为待插入链表和被插入链表，返回值小的头结点
        if pHead1.val < pHead2.val:
            pHead1.next = self.Merge(pHead1.next, pHead2)
            return pHead1
        else:
            pHead2.next = self.Merge(pHead1, pHead2.next)
            return pHead2