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删除有序链表中重复的元素-I

[编程题]删除有序链表中重复的元素-I

热度指数：162869 时间限制：C/C++ 1秒，其他语言2秒空间限制：C/C++ 256M，其他语言512M
算法知识视频讲解

删除给出链表中的重复元素（链表中元素从小到大有序），使链表中的所有元素都只出现一次
例如：
给出的链表为 $1\to1\to2$ ,返回 $1 \to 2$ .
给出的链表为 $1\to1\to 2 \to 3 \to 3$ ,返回 $1\to 2 \to 3$ .

数据范围：链表长度满足 $0 \le n \le 100$ ，链表中任意节点的值满足 $|val| \le 100$

进阶：空间复杂度 $O(1)$ ，时间复杂度 $O(n)$

示例1

输入

{1,1,2}

输出

{1,2}

示例2

输入

{}

输出

{}

说明：本题目包含复杂数据结构ListNode，点此查看相关信息

算法知识视频讲解

Python3

LeoAlan

class Solution:
    def deleteDuplicates(self , head: ListNode) -> ListNode:
        cur = head

        while cur and cur.next:
            if cur.val == cur.next.val:
                cur.next = cur.next.next
            else:
                cur = cur.next

        return head

发表于 2023-10-08 23:17:21 回复(0)

Python3

Honey、

#第一种

class Solution:

def deleteDuplicates(self , head: ListNode) -> ListNode:

# write code here

num = []

while head:

if head.val not in num:

num.append(head.val)

head = head.next

else:

head = head.next

new = ListNode(-1)

cur = new

for i in num:

cur.next = ListNode(i)

cur = cur.next

return new.next

#第二种

class Solution:

def deleteDuplicates(self , head: ListNode) -> ListNode:

# write code here

if head == None:

return head

pre = head

while pre and pre.next:

cur = pre.next

if cur.val == pre.val:

pre.next = cur.next

else:

pre = cur

return head

发表于 2023-06-10 18:41:55 回复(0)

Python3

这就开摆的小白很成熟

# 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:
# write code here
new_list = []
if head == None:
return head
while head:
new_list.append(head.val)
head = head.next
tmp = set(new_list)
a = list(tmp)
a.sort()
#print(a)
head = ListNode(0)
cur = head

for i in range(len(a)):
cur.next = ListNode(a[i])
cur = cur.next
return head.next

发表于 2023-03-23 12:18:29 回复(0)

Python3

牛客1217718号

class Solution:

def deleteDuplicates(self , head: ListNode) -> ListNode:

cur = head

while cur and cur.next:

nex = cur.next

if nex.val <= cur.val:

cur.next = nex.next

else:

cur = nex

return head

发表于 2022-09-30 21:51:48 回复(0)

Python3

牛客395635779号

class Solution:
def deleteDuplicates(self , head: ListNode) -> ListNode:
# write code here
cur = head
while cur:
while cur.next and cur.val == cur.next.val:
cur.next = cur.next.next
cur = cur.next
return head

发表于 2022-04-05 10:59:47 回复(0)