Given a sorted linked list, delete all duplicates such that each element appear onlyonce. For example, Given1->1->2, return1->2. Given1->1->2->3->3, return1->2->3. 初步构想:处理的对象是Sorted List,既然是已经排好序的链表,所以需要做的就是逐个比较,先将首元素加入到结果链表,然后遍...
Given a sorted linked list, delete all duplicates such that each element appear onlyonce. For example, Given1->1->2, return1->2. Given1->1->2->3->3, return1->2->3. 思路:遍历链表,通过两个指针保存当前位置和当前位置的前一个位置。假设两指针所指节点的值相等。则删除当前节点。假设不等。
本题类似于 LeetCode 82. Remove Duplicates from Sorted List II, 解题思路同 LeetCode: 82. Remove Duplicates from Sorted List II。只需要将删除当前节点的代码注释掉即可。 AC 代码 AI检测代码解析 /** * Definition for singly-linked list. * struct ListNode { * int v...
leetcode 83. Remove Duplicates from Sorted List Given a sorted linked list, delete all duplicates such that each element appear onlyonce. For example, Given1->1->2, return1->2. Given1->1->2->3->3, return1->2->3. # Definition for singly-linked list. # class ListNode(object): #...
Can you solve this real interview question? Remove Duplicates from Sorted Array - Given an integer array nums sorted in non-decreasing order, remove the duplicates in-place [https://en.wikipedia.org/wiki/In-place_algorithm] such that each unique element
Can you solve this real interview question? Remove Duplicates from Sorted Array - Given an integer array nums sorted in non-decreasing order, remove the duplicates in-place [https://en.wikipedia.org/wiki/In-place_algorithm] such that each unique element
- LeetCodeleetcode.com/problems/remove-duplicates-from-sorted-array/description/ 解题思路 双指针 class Solution { public: int removeDuplicates(vector<int>& nums) { if(nums.size() == 0) return 0; int i = 0, j = 1; while(j < nums.size()) { if(nums[i] != nums[j]) { i+...
26. 删除有序数组中的重复项 - 给你一个 非严格递增排列 的数组 nums ,请你 原地 [http://baike.baidu.com/item/%E5%8E%9F%E5%9C%B0%E7%AE%97%E6%B3%95] 删除重复出现的元素,使每个元素 只出现一次 ,返回删除后数组的新长度。元素的 相对顺序 应该保持 一致 。然后返回 n
📺 视频题解📖 文字题解 方法一:双指针这道题目的要求是:对给定的有序数组 \textit{nums} 删除重复元素,在删除重复元素之后,每个元素只出现一次,并返回新的长度,上述操作必须通过原地修改数组的方法,使用 O(1) 的空间复杂度完成。由于给定的数组 \textit{nums} 是有序的,因此对于任 双指针 Python JavaScrip...
增加一个count来计数,从左至右遍历,遇到相同的直接del,否则count += 1 classSolution:defremoveDuplicates(self,nums):""" :type nums: List[int] :rtype: int """count=0whilecount<len(nums)-1:ifnums[count]==nums[count+1]:# The same with nums.pop(count) or nums.remove(nums[count])delnums...