[英]How to find highest ,second highest number, Lowest Second Lowest number in given Array
任何人都可以請告訴我如何在給定數組中找到最高,第二高的數字,最低的第二低的數字
var numbers = new[] {855,3,64,6,24,75,3,6,24,45};
任何指針和建議都會非常有幫助。 謝謝
使用 Linq 概念
var a = new int[] { 855, 3, 64, 6, 24, 75, 3, 6, 24, 45 };
var Max = a.Max(z => z);
var Min = a.Min( z => z);
var SMax = a.OrderByDescending(z=>z).Skip(1).First();
var SMin = a.OrderBy(z => z).Skip(1).First();
您沒有指定復雜性要求:一種方法是按降序對數組進行排序,然后選擇頂部、第二個和第三個項目。
另一種是構建一個Heap,然后執行3次remove root(每次移除后重建堆)。
假設您在數組中至少有 2 個項目,您可以使用OrderBy()
和ElementAt()
:
var numbers = new[] { 855, 3, 64, 6, 24, 75, 3, 6, 24, 45 };
var secondLowest = numbers.OrderBy(num => num).ElementAt(1);
var secondHighest = numbers.OrderBy(num => num).Reverse().ElementAt(1);
獲取最高和最低值更簡單,可以使用Max()
和Min()
LINQ 方法完成。
var lowest = numbers.Min();
var highest = numbers.Max();
如果您擔心復雜性,則可以使用Selection algorithm獲得更好的結果。 使用它,您可以執行O(n)復雜度的操作。
你也可以試試這個——
using System;
using System.Collections.Generic;
using System.Linq;
using System.Web;
using System.Web.UI;
using System.Web.UI.WebControls;
public partial class secondHighestLowest : System.Web.UI.Page
{
int[] arr = new int[10] { 45, 3, 64, 6, 24, 75, 3, 6, 24, 45 };
protected void Page_Load(object sender, EventArgs e)
{
secondHighestLowestNumber();
secoundLowestNumber();
}
private void secondHighestLowestNumber()
{
int firstHighestNumber = arr[0];
int secondHighestNumber = arr[0];
for(int i = 0; i<arr.Length; i++)
{
if (arr[i]>firstHighestNumber)
{
firstHighestNumber = arr[i];
}
}
for (int x = 0; x < arr.Length; x++)
{
if (arr[x]>secondHighestNumber && firstHighestNumber!=arr[x])
{
secondHighestNumber = arr[x];
}
}
Response.Write("secondHighestNumber---- " + secondHighestNumber + "</br>");
}
private void secoundLowestNumber()
{
int firstLowestNumber = arr[0];
int secondLowestNumber = arr[0];
for (int i = 0; i < arr.Length; i++)
{
if (arr[i] < firstLowestNumber)
{
firstLowestNumber = arr[i];
}
}
for (int x = 0; x < arr.Length; x++)
{
if (arr[x] < secondLowestNumber && firstLowestNumber != arr[x])
{
secondLowestNumber = arr[x];
}
}
Response.Write("secondLowestNumber---- " + secondLowestNumber + "</br>");
}
}
希望這有幫助:)
int[] i = new int[] { 4, 8, 1, 9, 2, 7, 3 };
Array.Sort(i);
Console.WriteLine("Highest number :" + i[i.Length-1]);
Console.WriteLine("Second highest number :"+i[i.Length-2]);
Console.WriteLine("Lowest number :" + i[i.Length-i.Length]);
Console.WriteLine("Second Lowest number :" + i[i.Length -i.Length+1]);
Output : Highest number : 9
Second highest number : 8
Lowest number : 1
Second Lowest number : 2
為什么可以在其中完成兩個循環
int[] myArray = new int[] { 2, 4, 3, 6, 9 };
int max1 = 0;
int max2 = 0;
for (int i = 0; i < myArray.Length; i++)
{
if (myArray[i] > max1)
{
max2 = max1;
max1 = myArray[i];
}
else
{
max2 = myArray[i];
}
}
Console.WriteLine("first" + max1.ToString());
Console.WriteLine("Second" + max2.ToString());
Console.ReadKey();
我首先使用選擇排序算法按升序排列它們,然后顯示值。
static void Main(string[] args)
{
int[] Num =new int[] { 3, 4, 5, 6, 7, 0,99,105,55 };
int temp;
for(int a=0;a<Num.Length-1;a++)
{
for(int b=a+1;b<Num.Length;b++)
{
if(Num[a]>Num[b])
{
temp = Num[a];
Num[a] = Num[b];
Num[b] = temp;
}
}
}
Console.WriteLine("Max value ="+Num[Num.Length-1]+"\nSecond largest Max value="+ Num[Num.Length - 2]+"\nMin value =" + Num[0] + "\nSecond smallest Min value=" + Num[1]);
Console.WriteLine("\nPress to close");
Console.ReadLine();
}
即使有重復記錄,該算法也適用。
int[] intArrayInput= new int[] { 855, 3, 64, 6, 24, 75, 3, 6, 24, 45, 855 };
使用排序:
Array.Sort(intArrayInput);
int intMax = intArrayInput[intInput.Length - 1];
int intLow = intArrayInput[0];
並使用 LINQ:
int intSecondMax = intArrayInput.OrderByDescending(x => x).Distinct().Skip(1).First();
int intSecondLow = intArrayInput.OrderBy(x => x).Distinct().Skip(1).First();
Array.Sort(new); string newNumber = string.Join(" ", new[0]); return newNumber;
Array.Sort(new); string newNumber = string.Join(" ", new[1]); return newNumber;
Array.Sort(new); string newNumber = string.Join(" ", new[new.Length - 1]); return newNumber;
Array.Sort(new); string newNumber = string.Join(" ", new[new.Length - 1]); return newNumber;
使用常規方法找出最大、第二大、最小和第二小的數字的最短方法:
int LargestNumber = 0;
int SecondLargestNumber = 0;
int SmallestNumber = 0;
int SecondSmallestNumber = 0;
int[] array = new int[] { 250, 5, 17, 50, 98, 352, 2, 8, 67, 150, 200, 1 };
for (int i = 0; i < array.Length; i++)
{
if (array[i] > LargestNumber)
{
SecondLargestNumber = LargestNumber;
LargestNumber = array[i];
}
else if (array[i] > SecondLargestNumber)
{
SecondLargestNumber = array[i];
}
if (i == 0)
{
SmallestNumber = array[0];
SecondSmallestNumber = array[1];
}
if (array[i] < SmallestNumber)
{
SecondSmallestNumber = SmallestNumber;
SmallestNumber = array[i];
}
else if (array[i] < SecondSmallestNumber)
{
SecondSmallestNumber = array[i];
}
}
Console.WriteLine("Largest Number : " + LargestNumber);
Console.WriteLine("Second Largest Number : " + SecondLargestNumber);
Console.WriteLine("Smallest Number : " + SmallestNumber);
Console.WriteLine("Second Smallest Number : " + SecondSmallestNumber);
int[] myUnSortArray = { 1, 5, 8, 3, 10, 6, 19, 5, 4, 4 };
int[] SortedArray = (from number in myUnSortArray
orderby number ascending
select number).ToArray();
int highestValue = SortedArray.Max();
int SecondHighest = SortedArray.Last(m => m < highestValue);
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