场景 有一张得分表(score),记录了用户每次的得分,同一个人可能有多个得分. id name score 1 tom 45 2 jack 78 3 tom 34 . . . 需求:找出分数最高的前5个人. SQL1 首先我们写个最简单的sql: select id, name, score from score order by score desc limit 5; 如果sql这样写,结果可能是: id name score 2 jack 78 1 tom 45 3 tom 34 排序了,
举个栗子/**查询员工信息,要求工资按照从高到低进行排序(默认升序)**/SELECT * FROM employees ORDER BY salary ASC;/**方法2:**/SELECT * FROM employees ORDER BY salary;/**查询部门标号大于等于90的员工信息,并且按照入职时间进行先后排序**/SELECT * FROM employees WHERE department_id >= 90ORDER BY hiredata ASC;/**按照员工的年薪
//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
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
第一个题目: int a[] = {12,13,12,13,19,18,15,12,15,16,17},要求对数组a进行排序,要求时间复杂度为O(N) 我们所知道的常规排序中,最优的解法也就是O(N*log2^N),那如何做到时间复杂度为O(N)呢? 运用哈希算法的思想就可以优化算法为O(N) void Sort(int* a, int n) { assert(a); const int N = 20; int b[N] = { 0 }; for (int i = 0; i < n; i++)
[题目] 原文: 1.3 Design an algorithm and write code to remove the duplicate characters in a string without using any additional buffer. NOTE: One or two additional variables are fine. An extra copy of the array is not. FOLLOW UP Write the test cases for
NAME uniq - report or omit repeated lines SYNOPSIS uniq [OPTION]... [INPUT [OUTPUT]] DESCRIPTION Filter adjacent matching lines from INPUT (or standard input), writing to OUTPUT (or standard output). With no options, matching lines are merged to the