1 # -*- coding:utf-8 -*- 2 # class ListNode: 3 # def __init__(self, x): 4 # self.val = x 5 # self.next = None 6 """没有三个指针解决不了的问题""" 7 class Solution: 8 def deleteDuplication(self, pHead):##最好以原链表指针为输出 9 # write code here 10 pPre = None 11 pCur = pHead#给链表加个指针 12 pNext = None 13 while pCur != None: 14 if pCur.next != None and pCur.val == pCur.next.val: #如果不到尾节点，并且当前指针节点与下一节点数值相等时 15 pNext = pCur.next#更新next指针 16 while pNext.next != None and pNext.next.val == pCur.val: #当next指针没到尾节点，且next下一节点数值与当前节点值相等时 17 pNext = pNext.next #一直更新next指针 18 if pCur == pHead: #处理第一个节点就是重复节点的情况 19 pHead = pNext.next #直接将原链表pHead指针后移到一段重复的节点之后 20 else: 21 pPre.next = pNext.next #如果头结点不是重复节点，则移动pre指针 22 pCur = pNext.next #更新当前指针到一段重复节点之后节点 23 else: #更新前指针pre到现指针cur上 24 pPre = pCur 25 pCur = pCur.next #更新当前指针cur 26 return pHead