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
        dummy = ListNode(None)
        head = dummy
        cur1 = pHead1
        cur2 = pHead2
        while cur1 and cur2: # 对两个链表当前节点都存在的情况下，进行遍历赋值
            if cur1.val < cur2.val:
                dummy.next = cur1
                cur1 = cur1.next
            else:
                dummy.next = cur2
                cur2 = cur2.next
            dummy = dummy.next
        if not cur1:  # 有一个为空，下一个节点为另一个非空链表
            dummy.next = cur2 
        if not cur2:
            dummy.next = cur1 
        return head.next # 返回头节点的下一个节点