先说需求:找出一个对象List中,某个属性值最大的对象. 1.定义对象 private class A { public int ID { get; set; } public string Name { get; set; } } 2.为两种方法定义两个时间段全局变量. 1 private static TimeSpan compare = new TimeSpan(); private static TimeSpan order = new TimeSpan(); 3.第一种方法:对列表
找出最大值和最小值 题目要求 输入n个数,n<=100,找到其中最小的数和最大的数 实现代码 using System; namespace _1.求最大最小 { class Program { public static int GetMax(int[] numbers) { int max = numbers[0]; for (int i = 0; i < numbers.Length; i++) { if (max < numbers[i]) { max = numbers[i];
在一个SQL Server表中一行的多个列找出最大值 有时候我们需要从多个相同的列里(这些列的数据类型相同)找出最大的那个值,并显示 这里给出一个例子 IF (OBJECT_ID('tempdb..##TestTable') IS NOT NULL) DROP TABLE ##TestTable CREATE TABLE ##TestTable ( ID ,) PRIMARY KEY, Name ), UpdateByApp1Date DATETIME, UpdateByApp2Date DAT
# include <stdio.h> # define N main(){ int a, b; ,,,,,,,,,,,,,,,,}; //array中输入需要排序的数字 ]; ; a < N; a++){ if(array[a]>max){ max = array[a]; //使用max函数,快速筛选出最大值 } } printf("Max = %d\n",max,b); }
#输入若干个整数,打印出最大值 # m = int(input('Input first number >>>')) while True: c = input('Input a number >>>') if c: n = int(c) if n > m: m = n print('Max is',m) else: break
Given a string s and a non-empty string p, find all the start indices of p's anagrams in s. Strings consists of lowercase English letters only and the length of both strings s and p will not be larger than 20,100. The order of output does not matter.
Input: s: "abab" p: "ab" Output: [0, 1, 2] Explanation: The substring with start index = 0 is "ab", which is an anagram of "ab". The substring with start index = 1 is "ba", which is an anagram of "ab&
Given a string s and a non-empty string p, find all the start indices of p's anagrams in s. Strings consists of lowercase English letters only and the length of both strings s and p will not be larger than 20,100. The order of output does not matter.
有时候我们需要从多个相同的列里(这些列的数据类型相同)找出最大的那个值,并显示 这里给出一个例子 IF (OBJECT_ID('tempdb..##TestTable') IS NOT NULL) DROP TABLE ##TestTable CREATE TABLE ##TestTable ( ID INT IDENTITY(1,1) PRIMARY KEY, Name NVARCHAR(40), UpdateByApp1Date DATETIME, UpdateByApp2Date DATE
［抄题］: he set S originally contains numbers from 1 to n. But unfortunately, due to the data error, one of the numbers in the set got duplicated to another number in the set, which results in repetition of one number and loss of another number. Given a
