LeetCode–Remove_Duplicates_from_Sorted_List_II
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.
# _*_ coding: utf-8 _*_
class ListNode(object):
def __init__(self, x):
self.val = x
self.next = None
class Solution(object):
def deleteDuplicates(self, head):
"""
:type head: ListNode
:rtype: ListNode
"""
if not head or not head.next:
return head
dummy = ListNode(None)
node, dummy.next, last_uniq = head, head, dummy
last_val = None
while node:
# 先找出一串相等值
while node and last_val == node.val:
node = node.next
# 假如是最后一个值，那么直接break
if not node or not node.next:
last_uniq.next = node
break
# 更新最后的值
last_val = node.val
# 重新判断更新last_val后当前节点和下一个节点
if node.val != node.next.val:
last_uniq.next = node
last_uniq = last_uniq.next
node = node.next
return dummy.next
自己的提交虽然全部通过测试用例，但是leetcode评分为1.46％，感觉太差劲于是找到一个更好的简单的解法，贴在下面
# 以下为他人的解法
class Solution1(object):
'''算法思路：
每走一步判断当前节点是否应该被添加到列表里边
'''
def deleteDuplicates(self, head):
dummy = tail = ListNode(None)
while head:
node = head
# 这里他多判断了一次, 但是很灵活地跳出了所有重复的点
if head.next and head.next.val == node.val:
while head and head.val == node.val:
head = head.next
else:
tail.next, tail, head = head, head, head.next
tail.next = None
return dummy.next