计算值在数组中出现的次数

用于完成工作LINQ

using System.Linq;

var numQuery =
        from num in numbers
        where num == 5
        select num;

Console.WriteLine("Count of 5: " + numQuery.Count);

或者使用method syntax

var numQuery = numbers.Where(num => num == 5);
Console.WriteLine("Count of 5: " + numQuery.Count);

有关概述，请参阅此处，有关-syntax，请参阅此处。
找到 的样本，请看这里。query vs methodGroupBy

您的 count 数组有 4 个字段 ...

一个索引为 0、1、2 和 3

那么，如果像 4（或更大）这样的数字碰巧被计算在内会发生什么？YOR 代码尝试访问索引 4 ...不存在......

由于此部分，您收到索引越界错误：

for (i = 0; i < SIZE - 1; i++)
{
    if (numbers[i] > 0 && numbers[i] < SIZE)
    {
        x = Count[i];

请注意，当只有 .0SIZE - 111Count4

---

不过，使用 LINQ 可以很容易地完成此任务。

int[] numbers = new int[SIZE] { 5, 5, 5, 7, 7, 7, 9, 7, 9, 9, 9, 1 };

var count = numbers
    .GroupBy(e => e)
    .Where(e => e.Count() == 4)
    .Select(e => e.First());

因此，它按数字的值对数字进行分组，然后我们将列表细化为仅包含 4 组，然后选择每个组中的第一个，留下一个 s 的集合。int

---

下面是一个非基于 LINQ 的解决方案，它使用 Dictionary 来存储数字计数。

int[] numbers = new int[SIZE] { 5, 5, 5, 7, 7, 7, 9, 7, 9, 9, 9, 1 };
var dictionary = new Dictionary<int, int>();
var numbersWithFour = new List<int>();

foreach (var number in numbers)
{
    if (dictionary.ContainsKey(number))
        dictionary[number]++;
    else
        dictionary.Add(number, 1);
}

foreach (var val in dictionary)
{
    if (val.Value == 4)
    {
        numbersWithFour.Add(val.Key);
    }
}

---

通过对程序进行一些修改，您可以获得一些结果。

int[] numbers = new int[SIZE] { 5, 5, 5, 7, 7, 7, 9, 7, 9, 9, 9, 1 };
string[] letters = new string[SIZE] { "m", "m", "s", "m", "s", "s", "s", "m", "s", "s", "s", "s" };
int[] values = new int[SIZE] { 15, 22, 67, 45, 12, 21, 24, 51, 90, 60, 50, 44 };
string[] status = new string[SIZE] { "f", "m", "f", "a", "m", "f", "f", "f", "m", "f", "m", "f" };

// Set the size of Count to maximum value in numbers + 1
int[] Count = new int[9 + 1];
int x = 0;
int i = 0;

for (i = 0; i < SIZE - 1; i++)
{
    if (numbers[i] > 0 && numbers[i] < SIZE)
    {
        // Use value from numbers as the index for Count and increment the count
        Count[numbers[i]]++;
    }
}

for (i = 0; i < Count.Length; i++)
{
    // Check all values in Count, printing the ones where the count is 4
    if (Count[i] == 4)
        Console.WriteLine("{0}", i);
}

输出：

7
9

我使用正则表达式作为我的解决方案，因为我只有三个值。

String results = "" + one.ToString() + " " + two.ToString() + " " + three.ToString();
int count1 = Regex.Matches(results, @one.ToString()).Count;
int count2 = Regex.Matches(results, @two.ToString()).Count;
int count3 = Regex.Matches(results, @three.ToString()).Count;

看起来很“骇人听闻”，但对我有用。它可以处理字符串或数字，但前提是您正在处理几个值。在这种情况下非常有效。如果没有，我认为另一个答案会是一个更好的选择。

这是查找“计算值在数组中出现的次数”的幼稚解决方案 想法：在数组中构建哈希映射 溶液：

using System.Collections.Generic;
using System.Text;

namespace GetArrEleFrequency
{
    class Program
    {
      static int[] Arr = new int[5] { 3, 3, 0, 2, 0 };
      static int[] Key = new int[5];
      static int[] value = new int[5];
      static void Main(string[] args)
      {
         int keyItr = -1, ValueItr = -1, tempIndex = 0, tempValue = 0;
         for (int i=0; i <= Arr.Length-1;i++) {
              if (!(isPresent(Arr[i]))) {
                   keyItr += 1;ValueItr += 1;
                   Key[keyItr] = Arr[i];
                   value[ValueItr] = 1;
             } else {
                   
                   value[tempIndex] = value[getIndex(Arr[i])] + 1;
            }
        }
        for (int i=0;i<=Key.Length-1;i++) {
              Console.WriteLine(Key[i] + "-" + value[i]);
        }
        Console.ReadKey();
     }
     public static Boolean  isPresent(int num) {
      Boolean temp = false;
      for (int i=0; i <= Key.Length-1;i++) {
        if (Key[i] == num) {
              temp = true;
              break;
        } else {
              temp = false;
        }
     }
     return temp;
   }
   public static int getIndex(int num) {
       int temp = 0;
       for (int i=0;i<=Key.Length-1;i++) {
             if (Key[i] == num) {
               break;
             } else {
             temp += 1;
       }
  }
  return temp;
   }
 }
}

Output :

    3 - 2
    0 - 2
    2 - 1
    0 - 0
    0 - 0

 static void Main(string[] args)
        {
            int[] arr = new int[] { 45, 34, 23, 67, 10, 99,99,10 };
            foreach(int i in arr.Distinct())
            {
                int count = occurance(arr,i);
                Console.WriteLine(i + "-Occurred For :" + count);
            }
            
            Console.ReadLine();
        }
        public static int occurance(int[] arr,int x)
        {
            int count = 0;
            foreach(int num in arr)
            {
                if(x==num)
                {
                    count++;
                }
            }
            return count;
        }
    }

我认为如果不使用列表、LINQ 或字典，这个问题就无法得到回答，所以这是我的建议：

using System;
using System.Collections.Generic;
//using System.Linq;
using System.Text;

class Program
{
    static void Main()
    {
        int n = int.Parse(Console.ReadLine()); // the size of the array
        int[] ints = new int[n]; //an array to store the items, integers in this case
        int[] freq = new int[n]; //an array to store the frequency of each element with the same index

        for (int i = 0; i < n; i++) // a loop that takes each element on a new row
        {
            ints[i] = int.Parse(Console.ReadLine());
        }

        for (int j = 0; j < n; j++) // loops to iterate through the ints array and pick up the
                                    // frequencies and store them in the freq array
        {
            for (int k = 0; k < n; k++)
            {
                if (ints[j] == ints[k] && k != n)
                {
                    freq[j]++;
                }
            }
        }

        int indexAtMax = freq.ToList().IndexOf(freq.Max()); //this picks up the index of the first maximum count
        int mostFrequentNumber = ints[indexAtMax]; // the actual number behind the same inex in the ints array
        int frequencyOfRepeating = freq[indexAtMax]; // the actual number of the frequency 

        Console.WriteLine($"The most frequent number is:{mostFrequentNumber} and it repeats {frequencyOfRepeating} times)");
    }
}