[LeetCode] Remove Duplicates from Sorted List 移除有序链表中的重复项

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[LeetCode] Remove Duplicates from Sorted List 移除有序链表中的重复项

Given a sorted linked list, delete all duplicates such that each element appear only once. For example,Given 1->1->2, return 1->2.Given 1->1->2->3->3, return 1->2->3. 移除有序链表中的重复项需要定义个指针指向该链表的第一个元素,然后第一个元素和第二个元素比较,如果重复了,则删掉第二个元素,…

[LeetCode] 83. Remove Duplicates from Sorted List 移除有序链表中的重复项

Given a sorted linked list, delete all duplicates such that each element appear only once. Example 1: Input: 1->1->2 Output: 1->2 Example 2: Input: 1->1->2->3->3 Output: 1->2->3 这道题让我们移除给定有序链表的重复项,那么可以遍历这个链表,每个结点和其后面的结点比较,如果结点值相…

LeetCode 83. Remove Duplicates from Sorted List （从有序链表中去除重复项）

Given a sorted linked list, delete all duplicates such that each element appear only once. For example,Given 1->1->2, return 1->2.Given 1->1->2->3->3, return 1->2->3. 题目标签:Linked List 题目给了我们一个 list,让我们去除中间的重复项. 设一个 cursor = head…

【LeetCode每天一题】Remove Duplicates from Sorted List(移除有序链表中的重复数字)

Given a sorted linked list, delete all duplicates such that each element appear only once. Example 1: Input: 1->1->2 Output: 1->2 Example 2: Input: 1->1->2->3->3 Output: 1->2->3 思路对于链表类的题目主要考的是指针的操作,他需要对指针的指向节点有正确的操作.这道题我们可以使用从…

[CareerCup] 2.1 Remove Duplicates from Unsorted List 移除无序链表中的重复项

2.1 Write code to remove duplicates from an unsorted linked list.FOLLOW UPHow would you solve this problem if a temporary buffer is not allowed? 这道题让我们移除无序链表中的重复项,在LeetCode中有两道类似的题是Remove Duplicates from Sorted List 移除有序链表中的重复项和 Remove Duplicates fr…

[LeetCode] 80. Remove Duplicates from Sorted Array II ☆☆☆(从有序数组中删除重复项之二)

https://leetcode.com/problems/remove-duplicates-from-sorted-array-ii/discuss/27976/3-6-easy-lines-C%2B%2B-Java-Python-Ruby 描述 Follow up for "Remove Duplicates":What if duplicates are allowed at most twice? For example,Given sorted array A = [1,1…

LeetCode 83. Remove Duplicates from Sorted List（从有序链表中删除重复节点）

题意:从有序链表中删除重复节点. /** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ class Solution { public: ListNode* deleteDuplicates(ListNode* head) { if(head == NULL || head…