1、构造伪节点防止链表为空
2、充分利用递增排序的思想。从头节点开始比较

class Solution:
    # 返回合并后列表
    def Merge(self, pHead1, pHead2):
        dummy = ListNode(-1)
        cur = dummy
        while pHead1 and pHead2:
            if pHead1.val <= pHead2.val:
                cur.next = pHead1
                pHead1 = pHead1.next
            else:
                cur.next = pHead2
                pHead2 = pHead2.next
            cur = cur.next
        cur.next = pHead1 if pHead1 else pHead2
        return dummy.next