【python3】leetcode 82. Remove Duplicates from Sorted List II （Medium）_leetcode python deleteduplicates(self head):-优快云博客

本文链接：https://blog.youkuaiyun.com/maotianyi941005/article/details/86506201

本文介绍了一种算法，用于从已排序的链表中移除所有重复节点，仅保留原始列表中不重复的数字。通过两种方法实现：使用字典统计节点出现次数并遍历修改链表；在遍历时直接修改链表结构。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

82. Remove Duplicates from Sorted List II （Medium）

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

Example 1:
Input: 1->2->3->3->4->4->5
Output: 1->2->5
Example 2:
Input: 1->1->1->2->3
Output: 2->3

1 dict统计出现次数

# Definition for singly-linked list.
# 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
        node = head 
        dic = {}
        while(node):
            dic[node.val] = dic.get(node.val,0) + 1
            node = node.next
        pre = newhead = ListNode(0)
        newhead.next = head
        while(head):
            if dic[head.val] > 1:
                pre.next = head.next
            else:
                pre = head
               
            head = head.next
        return newhead.next

2 遍历时修改

用一个sameval 记录有和next的值相同的结点

class Solution(object):
    def deleteDuplicates(self, head):
        """
        :type head: ListNode
        :rtype: ListNode
        """
        if not head or not head.next:return head
        pre = newhead = ListNode(0)
        sameval = None
        while(head):
            if head.next: 
                if head.val == head.next.val:
                    sameval = head
                else:
                    if not sameval: # not duplicate
                        pre.next = head
                        pre = pre.next
                    sameval = None       
                head = head.next
            else:#lastone
                if not sameval:
                    pre.next = head
                else:pre.next=None
                break
                
        return newhead.next