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[LeetCode]82. 删除排序链表中的重复元素 II（Remove Duplicates from Sorted List II）

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题目描述

给定一个排序链表，删除所有含有重复数字的节点，只保留原始链表中 没有重复出现 的数字。

示例 1:

输入: 1->2->3->3->4->4->5
输出: 1->2->5

示例 2:

输入: 1->1->1->2->3
输出: 2->3

解题思路

  1. 解法一：递归。
  2. 解法二：双指针。慢指针指向已完成部分的最后一个节点，快指针指向当前节点，并完全跳过重复的节点。

代码

Python 3.6

解法一：递归

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

class Solution:
    def deleteDuplicates(self, head: ListNode) -> ListNode:
        if head is None or head.next is None:
            return head
        cur = head
        repeat = False
        while cur.next is not None and cur.val == cur.next.val:
            repeat = True
            cur = cur.next
        if repeat:
            return self.deleteDuplicates(cur.next)
        else:
            cur.next = self.deleteDuplicates(cur.next)
            return cur

执行用时 : 56 ms
内存消耗 : 13.7 MB

解法二：双指针

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

class Solution:
    def deleteDuplicates(self, head: ListNode) -> ListNode:
        if head is None or head.next is None:
            return head
        dummy = ListNode(head.val - 1)
        dummy.next = head
        slow = dummy
        fast = dummy.next
        while fast is not None:
            if fast.next is not None and fast.val == fast.next.val:
                tmp = fast.val
                while fast is not None and fast.val == tmp:
                    fast = fast.next
            else:
                slow.next = fast
                slow = slow.next
                fast = fast.next
        slow.next = fast
        return dummy.next

执行用时 : 56 ms
内存消耗 : 13.7 MB

参考

https://leetcode-cn.com/problems/remove-duplicates-from-sorted-list-ii/solution/kuai-man-zhi-zhen-by-powcai-2/

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