Question

我正在编写一个代码，用于打印出数组中重复的整数及其出现次数。我不允许使用LINQ，只是一个简单的代码。我想我是如此接近但对如何获得正确的输出感到困惑：

class Program
{
    static void Main(string[] args)
    {              
        int[] array = { 10, 5, 10, 2, 2, 3, 4, 5, 5, 6, 7, 8, 9, 11, 12, 12 };
        int count = 1;
        for (int i = 0; i < array.Length; i++)
        {
            for (int j = i; j < array.Length - 1 ; j++)
            {

               if(array[j] == array[j+1])
                  count = count + 1;
            }
            Console.WriteLine("\t\n " + array[i] + "occurse" + count);
            Console.ReadKey();
        }
    }
}

Answer 1

由于您无法使用LINQ，因此您可以使用集合和循环来执行此操作：

static void Main(string[] args)
{              
    int[] array = { 10, 5, 10, 2, 2, 3, 4, 5, 5, 6, 7, 8, 9, 11, 12, 12 };
    var dict = new Dictionary<int, int>();

    foreach(var value in array)
    {
        if (dict.ContainsKey(value))
            dict[value]++;
        else
            dict[value] = 1;
    }

    foreach(var pair in dict)
        Console.WriteLine("Value {0} occurred {1} times.", pair.Key, pair.Value);
    Console.ReadKey();
}

Answer 2

使用Group by:

int[] values = new []{1,2,3,4,5,4,4,3};

var groups = values.GroupBy(v => v);
foreach(var group in groups)
    Console.WriteLine("Value {0} has {1} items", group.Key, group.Count());

Answer 3

让我们看一个更简单的例子。假设我们有数组{0, 0, 0, 0}。

您的代码会做什么？

首先会查看第一项之后的项目数等于它。第一个之后有三个项目等于它。

然后它转到下一个项目，并在其后面查找与其相等的所有项目。那里有两个。到目前为止我们已经5岁了，我们还没有完成（我们还有一个要添加），但整个阵列中只有四个项目。

显然我们在这里有一个问题。我们需要确保当我们在数组中搜索给定项目的副本时，我们不再为同一项目搜索它。虽然有办法做到这一点，但这种基本方法看起来是相当多的工作。

当然，我们可以采取不同的方法。相反，遍历每个项目并搜索其他类似项目，我们可以遍历数组一次，并添加到我们找到该字符的次数计数。使用Dictionary可以轻松实现：

var dictionary = new Dictionary<int, int>();

foreach (int n in array)
{
    if (!dictionary.ContainsKey(n))
        dictionary[n] = 0;
    dictionary[n]++;
}

现在我们可以遍历字典并查看多次找到哪些值：

foreach(var pair in dictionary)
    if(pair.Value > 1)
        Console.WriteLine(pair.Key);

这使代码清晰可读，显然正确，并且（作为奖励）比代码更有效，因为您可以避免多次循环遍历集合。

Answer 4

好的我已经修改了你的代码。这应该做的工作：

class Program
{
    static void Main(string[] args)
    {
        int[] array = { 10, 5, 10, 2, 2, 3, 4, 5, 5, 6, 7, 8, 9, 11, 12, 12 };

        for (int i = 0; i < array.Length; i++)
        {
            int count = 0;
            for (int j = 0; j < array.Length; j++)
            {

                if (array[i] == array[j])
                    count = count + 1;
            }
            Console.WriteLine("\t\n " + array[i] + " occurs " + count + " times");
        }
        Console.ReadKey();
    }
}

Answer 5

这是一个避免使用词典的答案。由于OP说他不熟悉它们，这可能会让他对词典的作用有所了解。

这个答案的缺点是你必须对数组中的最大数量加强限制，而你不能有负数。你真的不会在实际代码中使用这个版本。

int[] array = { 10, 5, 10, 2, 2, 3, 4, 5, 5, 6, 7, 8, 9, 11, 12, 12 };
int[] count = new int[13];

foreach(int number in array) {
    // using the index of count same way you'd use a key in a dictionary
    count[number]++;
}

foreach(int c in count) {
    int numberCount = count[c];
    if(numberCount > 0) {
        Console.WriteLine(c + " occurs " + numberCount + " times");
    }
}

Answer 6

你犯了一个小错误，就是用J代替我......

class Program
{
    static void Main(string[] args)
    {              
        int[] array = { 10, 5, 10, 2, 2, 3, 4, 5, 5, 6, 7, 8, 9, 11, 12, 12 };
        int count = 1;
        for (int i = 0; i < array.Length; i++)
        {
            for (int j = i; j < array.Length - 1 ; j++)
            {
               if(array[i] == array[j+1])
                  count = count + 1;
            }
            Console.WriteLine("\t\n " + array[i] + "occurse" + count);
            Console.ReadKey();
        }
    }
}

Answer 7

int[] arr = { 10, 5, 10, 2, 2, 3, 4, 5, 5, 6, 7, 8, 9, 11, 12, 12 };
var result = arr.GroupBy(x => x).Select(x => new { key = x.Key, val = x.Count() });       
foreach (var item in result)
{
    if(item.val > 1)
    {                
        Console.WriteLine("Duplicate value : {0}", item.key);
        Console.WriteLine("MaxCount : {0}", item.val);
    }

}

Console.ReadLine();

Answer 8

int[] array = { 10, 5, 10, 2, 2, 3, 4, 5, 5, 7, 7, 8, 9, 7, 12, 12 };
Dictionary<int, int> duplicateNumbers = new Dictionary<int, int>();
int count=1;
for (int i = 0; i < array.Length; i++)
{
    count=1;
    if(!duplicateNumbers.ContainsKey(array[i]))
    {
        for (int j = i; j < array.Length-1; j++)
        {
            if (array[i] == array[j+1])
            {
                count++;                            
            }
        }
        if (count > 1)
        {
            duplicateNumbers.Add(array[i], count);
        }
    }
}
foreach (var num in duplicateNumbers)
{
    Console.WriteLine("Duplicate numbers, NUMBER-{0}, OCCURRENCE- {1}",num.Key,num.Value);
}

Answer 9

/ 这是帮助您使用Forloop查找重复整数值的答案，它将仅返回重复值而不是其出现次数 /

    public static void Main(string[] args)
    {
        //Array list to store all the duplicate values
        int[] ary = { 10, 5, 10, 2, 2, 3, 4, 5, 5, 6, 7, 8, 9, 11, 12, 12 };
        ArrayList dup = new ArrayList();

        for (int i = 0; i < ary.Length; i++)
        {
            for (int j = i + 1; j < ary.Length; j++)
            {
                if (ary[i].Equals(ary[j]))
                {
                    if (!dup.Contains(ary[i]))
                    {
                        dup.Add(ary[i]);
                    }
                }
            }
        }
        Console.WriteLine("The numbers which duplicates are");
        DisplayArray(dup);
    }
    public static void DisplayArray(ArrayList ary)
    {
        //loop through all the elements
        for (int i = 0; i < ary.Count; i++)
        {
            Console.Write(ary[i] + " ");
        }
        Console.WriteLine();
        Console.ReadKey();
    }

Answer 10

//此方法计算数组中重复元素的总数

public void DuplicateElementsInArray(int[] numbers)
    {
        int count = 0;

        for (int i = 0; i < numbers.Length; i++)
        {

            for (int j = i; j < numbers.Length - 1; j++)
            {
                if (numbers[i] == numbers[j+1])
                {                       
                    count++;
                    break;
                }                       
            }               
        }
        Console.WriteLine("Total number of duplicate elements found in the array is: {0}", count);
    }

Answer 11

static void printRepeating（int [] arr，整数大小） { 我

    Console.Write("The repeating" +  
                   " elements are : "); 

    for (i = 0; i < size; i++) 
    { 
        if (arr[ Math.Abs(arr[i])] >= 0) 
            arr[ Math.Abs(arr[i])] = 
                -arr[ Math.Abs(arr[i])]; 
        else
            Console.Write(Math.Abs(arr[i]) + " "); 
    }          
}

Answer 12

using System;
using System.Collections.Generic;

namespace ConsoleApp1
{
    /// <summary>
    /// How do you find the duplicate number on a given integer array?
    /// </summary>
    class Program
    {
        static void Main(string[] args)
        {
            int[] array = { 10, 5, 10, 2, 2, 3, 4, 5, 5, 6, 7, 8, 9, 11, 12, 12 };

            Dictionary<int, int> duplicates = FindDuplicate(array);

            Display(duplicates);

            Console.ReadLine();
        }

        private static Dictionary<T, int> FindDuplicate<T>(IEnumerable<T> source)
        {
            HashSet<T> set = new HashSet<T>();
            Dictionary<T, int> duplicates = new Dictionary<T, int>();

            foreach (var item in source)
            {
                if (!set.Add(item))
                {
                    if (duplicates.ContainsKey(item))
                    {
                        duplicates[item]++;
                    }
                    else
                    {
                        duplicates.Add(item, 2);
                    }
                }
            }

            return duplicates;
        }

        private static void Display(Dictionary<int, int> duplicates)
        {
            foreach (var item in duplicates)
            {
                Console.WriteLine($"{item.Key}:{item.Value}");
            }
        }
    }
}

Answer 13

 class Program
{
    static void Main(string[] args)
    {
        int[] arr = { 2, 3, 2, 4, 5, 12, 2, 3, 3, 3, 12 };
        List<int> nums = new List<int>();
        List<int> count = new List<int>();
        nums.Add(arr[0]);
        count.Add(1);
        for (int i = 1; i < arr.Length; i++)
        {

            if(nums.Contains(arr[i]))
            {
                count[nums.IndexOf(arr[i])] += 1;
            }
            else
            {
                nums.Add(arr[i]);
                count.Add(1);
            }
        }

        for(int x =0; x<nums.Count;x++)
        {
            Console.WriteLine("number:"+nums[x] +"Count :"+ count[x]);
        }
        Console.Read();
    }
}

Answer 14

int copt = 1;
int element = 0;
int[] array = { 1, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 5 };
for (int i = 0; i < array.Length; i++)
{
   for (int j = i + 1; j < array.Length - 1; j++)
   {
       if (array[i] == array[j])
       {
          element = array[i];
          copt++;
          break;
       }
   }
}

Console.WriteLine("the repeat element is {0} and it's appears {1} times ", element, copt);
Console.ReadKey();

//输出的元素是3，出现9次

Answer 15

 class Program
{
    static void Main(string[] args)
    {
        int[] arr = new int[] { 10, 20,20, 30, 10, 50 ,50,9};
        List<int> listadd = new List<int>();

        for (int i=0; i <arr.Length;i++)
        {
           int count = 0;
            int flag = 0;

            for(int j=0; j < arr.Length; j++)
            {
                if (listadd.Contains(arr[i]) == false)
                {


                    if (arr[i] == arr[j])
                    {
                        count++;
                    }

                } 

                else
                {
                    flag = 1;
                }

            }
            listadd.Add(arr[i]);
            if(flag!=1)
            Console.WriteLine("No of occurance {0} \t{1}", arr[i], count);
        }
        Console.ReadLine();

    }
}

Answer 16

public class Program
{
    public static void Main(string[] args)
    {
        int[] arr = new int[]{10, 5, 10, 2, 2, 3, 4, 5, 5, 6, 7, 8, 9, 11, 12, 12};

        int numberOfDuplicate = 0;
        int arrayLength = arr.Length;

        for(int i=0; i < arrayLength; i++)  
        {  
           for(int j = 1; j < arrayLength; j++)  
           {
               if(j < i && arr[j]==arr[i])
               {
                  break; 
               }
               else if(i != j && arr[i]==arr[j])
               {
                  Console.Write("duplicate : arr[{0}]={1}--arr[{2}]={3}\n",i,arr[i],j,arr[j]);
                  numberOfDuplicate++;
                  break;
               }
           }
        }

        Console.Write("Total duplicate element in an array 'arr' is {0}\n ",numberOfDuplicate);
    }
}

Answer 17

只需使用下面的代码，无需使用字典或ArrayList.Hope这有助于：）

public static void Main(string[] args)
{              
    int [] array =new int[]{10, 5, 10, 2, 2, 3, 4, 5, 5, 6, 7, 8, 9, 11, 12, 12};
    Array.Sort(array);
    for (int i = 0; i < array.Length; i++)
    {
        int count = 1,k=-1;
        for (int j = i+1; j < array.Length; j++)
        {  

                if(array[i] == array[j])
                {
                    count++;
                    k=j-1;
                }
                else break;
        }
        Console.WriteLine("\t\n " + array[i] + "---------->" + count);            
        if(k!=-1)
            i=k+1;          
    }
}

Answer 18

public static void Main（string [] args）

{

Int [] array = {10,5,10,2,3,3,5,5,6,7,8,9,11,12,12};

List<int> doneNumbers = new List<int>();


for (int i = 0; i < array.Length - 1; i++)

{

    if(!doneNumbers.Contains(array[i]))

    {

        int currentNumber = array[i];

        int count = 0;

        for (int j = i; j < array.Length; j++)

        {

            if(currentNumber == array[j])

            {

                count++;

            }

        }

        Console.WriteLine("\t\n " + currentNumber +" "+ " occurs " + " "+count + " "+" times");

        doneNumbers.Add(currentNumber);

        Console.ReadKey();
      }

   }

}

}

Answer 19

我同意使用Dictionary比嵌套for循环更好的运行时性能（O（n）vs O（n ^ 2））。但是，为了解决OP，这里有一个解决方案，其中HashSet用于防止重复计数已经计数的整数，例如示例数组中的整数5。

static void Main(string[] args)
{              
    int[] A = { 10, 5, 10, 2, 2, 3, 4, 5, 5, 6, 7, 8, 9, 11, 12, 12 };

    var set = new HashSet<int>();
    for (int i = 0; i < A.Length - 1; i++) {
        int count = 0;
        for (int j = i; j < A.Length - 1; j++) {
            if (A[i] == A[j + 1] && !set.Contains(A[i]))
                count++;
        }
        set.Add(A[i]);
        if (count > 0) {
            Console.WriteLine("{0} occurs {1} times", A[i], count + 1);
            Console.ReadKey();
        }
    }
}

Answer 20

这种修正的方法会给出正确的输出（效率非常低，但除非你大幅度扩大，否则这不是问题。）

int[] array = { 10, 5, 10, 2, 2, 3, 4, 5, 5, 6, 7, 8, 9, 11, 12, 12 };
for (int i = 0; i < array.Length; i++)
{
    int count = 0;
    for (int j = 0; j < array.Length ; j++)
    {
       if(array[i] == array[j])
          count = count + 1;
    }
    Console.WriteLine("\t\n " + array[i] + " occurs " + count);
    Console.ReadKey();
}

我在OP代码中计算了5个错误，如下所示。

int[] array = { 10, 5, 10, 2, 2, 3, 4, 5, 5, 6, 7, 8, 9, 11, 12, 12 };
int count = 1;                                   // 1. have to put "count" in the inner loop so it gets reset
                                                 // 2. have to start count at 0
for (int i = 0; i < array.Length; i++)
{
    for (int j = i; j < array.Length - 1 ; j++)  // 3. have to cover the entire loop
                                                 // for (int j=0 ; j<array.Length ; j++)
    {
       if(array[j] == array[j+1])                // 4. compare outer to inner loop values
                                                 // if (array[i] == array[j])
          count = count + 1;
    }
    Console.WriteLine("\t\n " + array[i] + "occurse" + count);
                                                 // 5. It's spelled "occurs" :)
    Console.ReadKey();
}

修改

要获得更好的方法，请使用Dictionary来跟踪计数。这允许您只循环一次数组，并且不会向控制台打印重复计数。

var counts = new Dictionary<int, int>();
int[] array = { 10, 5, 10, 2, 2, 3, 4, 5, 5, 6, 7, 8, 9, 11, 12, 12 };
for (int i = 0; i < array.Length; i++)
{
    int currentVal = array[i];
    if (counts.ContainsKey(currentVal))
        counts[currentVal]++;
    else
        counts[currentVal] = 1;
}
foreach (var kvp in counts)
    Console.WriteLine("\t\n " + kvp.Key + " occurs " + kvp.Value);

Answer 21

使用ToLookup：

int[] array = { 10, 5, 10, 2, 2, 3, 4, 5, 5, 6, 7, 8, 9, 11, 12, 12 };
var result = values.ToLookup(v => v).Select(x=>new {Value=x.Key,Count=x.Count() });

ToLookup返回允许索引的数据结构。它是一种扩展方法。我们得到一个ILookup实例，该实例可以使用foreach循环进行索引或枚举。这些条目在每个键处组合在一起。

Answer 22

您可以查看以下代码。代码正在运作。

 class Program {
    static void Main(string[] args) {
      int[] array = { 10, 5, 10, 2, 2, 3, 4, 5, 5, 6, 7, 8, 9, 11, 12, 12 };
      int count = 1;
      for (int i = 0; i < array.Length; i++) {
        for (int j = i + 1; j < array.Length; j++) {

          if (array[j] == array[j])
            count = count + 1;
        }
        Console.WriteLine("\t\n " + array[i] + " out of " + count);
        Console.ReadKey();
      }
    }
  }

在数组中查找重复的整数并显示它们出现的次数

22 个答案: