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
        flag = ListNode(-1)
        result = None
        # 如果两个链表都是空链表。
        if (not pHead1) and (not pHead2):
            return pHead1
        # 选择小的元素放入结果链表。
        while pHead1 and pHead2:
            if pHead1.val < pHead2.val:
                if result is None:
                    result = pHead1
                    # 记录结果链表的头节点。
                    flag.next = result
                    pHead1 = pHead1.next
                else:
                    result.next = pHead1
                    result = result.next
                    pHead1 = pHead1.next
            else:
                if result is None:
                    result = pHead2
                    flag.next = result
                    pHead2 = pHead2.next
                else:
                    result.next = pHead2
                    result = result.next
                    pHead2 = pHead2.next
        # 当一个链表的元素已经全部访问，直接把另一个链表的剩余部分接入到结果链表。
        if pHead1:
            # 如果有一个链表是空链表，没有经过上面的循环过程，此处也需要记录结果链表的头节点。
            if result is None:
                result = pHead1
                # 记录结果链表的头节点。
                flag.next = result
            else:
                result.next = pHead1

        else:
            if result is None:
                result = pHead2
                flag.next = result
            else:
                result.next = pHead2

        return flag.next