如何找到最高，人數名列第二，在給定Array最低次低數

Question

誰能請告訴我如何如何找到在給定Array

var numbers = new[] {855,3,64,6,24,75,3,6,24,45}; 

任何最高，人數名列第二，最低次低數指針和建議將非常有幫助。謝謝

Answer 1

您也可以嘗試這個 -

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>"); 
    } 
} 

希望這是有益:)

Answer 2

您沒有指定複雜性要求：一種方法是按降序對數組進行排序並選擇第一項，第二項和第三項。

另一個是建立一個堆，然後執行刪除根3次（每次刪除後重建堆）。

Answer 3

假設你已經在陣列中至少2項，你可以使用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); 

獲取最高和最低是簡單的，並且可以使用Max()和Min()完成 LINQ方法。

var lowest = numbers.Min(); 
var highest = numbers.Max(); 

如果你擔心的複雜性，你可以使用Selection algorithm達到更好的效果。使用它你可以執行O（n）複雜的操作。

Answer 4

 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 

Answer 5

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); 

Answer 6

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(); 

Answer 7

時，爲什麼可以做兩個循環在

 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(); 

Answer 8

我首先使用選擇排序算法按升序排列它們，然後我顯示值。

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(); 
    } 

Answer 9

該算法適用於重複記錄。

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]; 

而且使用LINQ：

int intSecondMax = intArrayInput.OrderByDescending(x => x).Distinct().Skip(1).First(); 
int intSecondLow = intArrayInput.OrderBy(x => x).Distinct().Skip(1).First();