剑指offer——16合并两个链表

class Solution:
    # 返回合并后列表
    def Merge(self, pHead1, pHead2):
        # write code here
        z = ListNode(90)
        p = z#定义头结点位置
        if pHead1==None:
            return pHead2
        if pHead2==None:
            return pHead1
        #当两个链表都有值时
        while pHead1 and pHead2:
            #如果p1的值大于p2的值，p2的值应该进行头插
            if pHead1.val>=pHead2.val:
                z.next = pHead2
                pHead2 = pHead2.next
            #如果p1的值小于p2的值，p1后移
            else:
                z.next = pHead1
                pHead1 = pHead1.next
            z = z.next
        if pHead1:
            z.next = pHead1
        if pHead2:
            z.next = pHead2
        
        return p.next