//定义一个100元素的集合,包含A-Z List<String> list = new LinkedList<>(); for (int i =0;i<100;i++){ list.add(String.valueOf((char)('A'+Math.random()*('Z'-'A'+1)))); } System.out.println(list); //统计集合重复元素出现次数,并且去重返回hashmap Map<String, Long> map = l
分别使用List中Distinct(),GroupBy()实现链表的去重. 1.先上效果: 一维链表中分别有元素“aa”,"bb",'aa','aa',"cc",使用Distinct()方法后输出 aa,bb,cc 二维链表中类型为ClassA类型,其中对象的属性A分别为1,1,2,3,1,使用GroupBy()方法实则是分类,输出Key值分别为1,2,3. 2.上代码,类ClassA class ClassA { private int a; public int
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 题意: 有序链表去重 思路: 代码: class Solution { public ListNod
Given a sorted linked list, delete all nodes that have duplicate numbers, leaving only distinct numbers from the original list. For example,Given 1->2->3->3->4->4->5, return 1->2->5.Given 1->1->1->2->3, return 2->3.
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. 简单的链表去重而已啊,遍历一边就实现了: class Solution { public: ListNode* delet
题目: 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. 说明:给定已排好序的单链表,去重! 1)此链表默认无头节点,head指针指向的是第一个节点 2)链表为空或者只有一