[LeetCode] 82. Remove Duplicates from Sorted List II 从分类列表中删除重复项目 II @python

Description

Given a sorted linked list, delete all nodes that have duplicate numbers, leaving only distinct numbers from the original list.

For example,
Given 1->2->3->3->4->4->5, return 1->2->5.
Given 1->1->1->2->3, return 2->3.

给定一个有序的链表，删除所有有重复数字的节点，只保留原始列表中唯一的数字。

例如：
给定 1->2->3->3->4->4->5 ，则返回 1->2->5
给定 1->1->1->2->3 ，则返回 2->3

Solution

本题使用双指针，pre 和 cur， pre指向表头。只有当pre的指向与cur的指向不同时，pre才移动到cur的指向，碰到重复的项目，则cur移动一位，直到两个指针指向不同。注意当

pre.next == cur

的情况。

# -*- coding: utf-8 -*-
"""
Created on Sun Mar 18 10:10:07 2018

@author: Saul
"""
# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
def deleteDuplicates(self, head):
"""
:type head: ListNode
:rtype: ListNode
"""

if head == None or head.next == None:
return head

dummy = ListNode(0)
dummy.next = head
pre = dummy
cur = head
while cur:
while cur.next and pre.next.val == cur.next.val:
cur = cur.next
if pre.next == cur:   # think about this
pre = pre.next
else:
pre.next = cur.next  # and this
cur = cur.next
return dummy.next