冒泡排序 // 冒泡排序 bubble sort public static int[] BubbleSort(int []array) { bool isContinue = true; ; i < array.Length && isContinue; i++) { isContinue = false; ; j < array.Length - i - ; j++) { ]) { int temp = array[j]; array[j] = array[j + ]; a…

1: 找小于最大的最大的 select max(Salary) from Employee where Salary<(select MAX(Salary) from Employee); 2. 排序 select Salary from Employee where Salary not in (select MAX(Salary)from Employee) order by Salary desc limit 1; select ( select distinct Salary from…

装载声明:http://blog.csdn.net/lxsmk9059/article/details/77920206?locationNum=1&fps=1 ,,,,,,,,}; ]; ]; ; i < sizeof(array)/sizeof(int); i++) { if(array[i] > max) { secondmax = max; max = array[i]; } else if(array[i] > secondmax) { secondmax = arra…

, NULL, salary) as `salary` from ( ,) tmp Write a SQL query to get the second highest salary from the Employee table. +----+--------+ | Id | Salary | +----+--------+ | 1 | 100 | | 2 | 200 | | 3 | 300 | +----+--------+ For example, given the above Emp…

#!usr/bin/env python #encoding:utf-8 ''''' __Author__:沂水寒城 功能:找出数组中第2大的数字 ''' def find_Second_large_num(num_list):   '''''   找出数组中第2大的数字   '''   #直接排序,输出倒数第二个数即可   tmp_list=sorted(num_list)   print 'Second_large_num is:', tmp_list[-2]   #设置两个标志位一个存储最…

using System; using System.Collections.Generic; using System.Linq; using System.Text; using System.Threading.Tasks; namespace ConsoleApplication9 { class Program { static void Main(string[] args) { // sql如下,我把它翻译成代码,加深理解 // SELECT MAX(id)FROM sys_men…

#include<stdio.h> int main() { int n,m,t,max,max1; scanf("%d",&n); while(n--) { max=-; max1=-; scanf("%d",&m); while(m--) { scanf("%d",&t); if(t==max || t==max1) continue; if(t>max) { max1=max; max=t; } e…

Write a SQL query to get the second highest salary from the Employee table. +----+--------+ | Id | Salary | +----+--------+ | 1 | 100 | | 2 | 200 | | 3 | 300 | +----+--------+ For example, given the above Employee table, the second highest salary is …

找第二大 # Write your MySQL query statement below SELECT MAX(Salary) FROM Employee WHERE Salary NOT IN (SELECT MAX(Salary) FROM Employee)…

问题:找出一个数组里面前K个最大数. 解法一(直接解法): 对数组用快速排序,然后直接挑出第k大的数.这种方法的时间复杂度是O(Nlog(N)).N为原数组长度. 这个解法含有很多冗余,因为把整个数组都排序了,而实际上我们不需要这样做. 解法二(K数组排序): 首先,创建一个长度为K的空数组.从原数组中先挑出K个数进行排序并放到这个空数组中.这一步的时间复杂度是O(Klog(K)). 接着,从剩下的N-K个值中,依次遍历并与上面数组的末尾的数(即里面的最大数)相比较,并插入到合适位置,需要执行(…