题解 | #删除有序链表中重复的元素-I#

# 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 head ListNode类 
# @return ListNode类
#
class Solution:
    def deleteDuplicates(self , head: ListNode) ->ListNode:
        #判断链表是否为空
		if head == None:
            return None
		  #遍历链表 转化为数组
        p = head
        nums = []
        while p != None :
                nums.append(p.val)
                p = p.next
        new1 = []
		#遍历数组去重
        for i in nums:
            if i not in new1:
                new1.append(i)
		#构建虚拟节点 将数组转为链表
        p1 = ListNode(-1)
        p2 = p1
        for i in new1:
            temp = ListNode(i)
            p2.next = temp
            p2 = p2.next
        return p1.next