找到重复的和丢失的|添加了3种新方法

给定大小为n的未排序数组。数组元素的范围是1到n。集合{1，2，…n}中的一个数字丢失，并且一个数字在数组中出现两次。找到这两个数字。

例子：

Input: arr[] = {3, 1, 3}
Output: Missing = 2, Repeating = 3
Explanation: In the array, 
2 is missing and 3 occurs twice 

Input: arr[] = {4, 3, 6, 2, 1, 1}
Output: Missing = 5, Repeating = 1

以下是解决问题的各种方法：

方法1(使用排序)
方法：

• 对输入数组进行排序。
• 遍历数组并检查是否丢失和重复。

时间复杂度： O(nLogn)

感谢LoneShadow提出了这种方法。

方法2(使用计数数组)
方法：

• 创建一个大小为n的临时数组temp []，所有初始值均为0。
• 遍历输入数组arr []，并对每个arr [i]执行以下操作
  • if(temp [arr [i]] == 0)temp [arr [i]] = 1;
  • if(temp [arr [i]] == 1)输出“ arr [i]” //重复
• 遍历temp []并输出值为0的数组元素(这是缺少的元素)

时间复杂度： O(n)

辅助空间： O(n)

方法3(使用元素作为索引并标记访问的地方)
方法：
遍历数组。遍历时，将每个元素的绝对值用作索引，并使该索引处的值为负，以将其标记为已访问。如果某些内容已标记为否，则为重复元素。要查找丢失的内容，请再次遍历数组并寻找一个正值。

C++

// C++ program to Find the repeating
// and missing elements
  
#include 
using namespace std;
  
void printTwoElements(int arr[], int size)
{
    int i;
    cout << " The repeating element is ";
  
    for (i = 0; i < size; i++) {
        if (arr[abs(arr[i]) - 1] > 0)
            arr[abs(arr[i]) - 1] = -arr[abs(arr[i]) - 1];
        else
            cout << abs(arr[i]) << "\n";
    }
  
    cout << "and the missing element is ";
    for (i = 0; i < size; i++) {
        if (arr[i] > 0)
            cout << (i + 1);
    }
}
  
/* Driver code */
int main()
{
    int arr[] = { 7, 3, 4, 5, 5, 6, 2 };
    int n = sizeof(arr) / sizeof(arr[0]);
    printTwoElements(arr, n);
}
  
// This code is contributed by Shivi_Aggarwal

C

// C program to Find the repeating
// and missing elements
  
#include 
#include 
  
void printTwoElements(int arr[], int size)
{
    int i;
    printf("\n The repeating element is");
  
    for (i = 0; i < size; i++) {
        if (arr[abs(arr[i]) - 1] > 0)
            arr[abs(arr[i]) - 1] = -arr[abs(arr[i]) - 1];
        else
            printf(" %d ", abs(arr[i]));
    }
  
    printf("\nand the missing element is ");
    for (i = 0; i < size; i++) {
        if (arr[i] > 0)
            printf("%d", i + 1);
    }
}
  
// Driver code
int main()
{
    int arr[] = { 7, 3, 4, 5, 5, 6, 2 };
    int n = sizeof(arr) / sizeof(arr[0]);
    printTwoElements(arr, n);
    return 0;
}

Java

// Java program to Find the repeating
// and missing elements
  
import java.io.*;
  
class GFG {
  
    static void printTwoElements(int arr[], int size)
    {
        int i;
        System.out.print("The repeating element is ");
  
        for (i = 0; i < size; i++) {
            int abs_val = Math.abs(arr[i]);
            if (arr[abs_val - 1] > 0)
                arr[abs_val - 1] = -arr[abs_val - 1];
            else
                System.out.println(abs_val);
        }
  
        System.out.print("And the missing element is ");
        for (i = 0; i < size; i++) {
            if (arr[i] > 0)
                System.out.println(i + 1);
        }
    }
  
    // Driver code
    public static void main(String[] args)
    {
        int arr[] = { 7, 3, 4, 5, 5, 6, 2 };
        int n = arr.length;
        printTwoElements(arr, n);
    }
}
  
// This code is contributed by Gitanjali

Python3

# Python3 code to Find the repeating 
# and the missing elements
  
def printTwoElements( arr, size):
    for i in range(size):
        if arr[abs(arr[i])-1] > 0:
            arr[abs(arr[i])-1] = -arr[abs(arr[i])-1]
        else:
            print("The repeating element is", abs(arr[i]))
              
    for i in range(size):
        if arr[i]>0:
            print("and the missing element is", i + 1)
  
# Driver program to test above function */
arr = [7, 3, 4, 5, 5, 6, 2]
n = len(arr)
printTwoElements(arr, n)
  
# This code is contributed by "Abhishek Sharma 44"

C#

// C# program to Find the repeating
// and missing elements
  
using System;
  
class GFG {
    static void printTwoElements(int[] arr, int size)
    {
        int i;
        Console.Write("The repeating element is ");
  
        for (i = 0; i < size; i++) {
            int abs_val = Math.Abs(arr[i]);
            if (arr[abs_val - 1] > 0)
                arr[abs_val - 1] = -arr[abs_val - 1];
            else
                Console.WriteLine(abs_val);
        }
  
        Console.Write("And the missing element is ");
        for (i = 0; i < size; i++) {
            if (arr[i] > 0)
                Console.WriteLine(i + 1);
        }
    }
  
    // Driver program
    public static void Main()
    {
        int[] arr = { 7, 3, 4, 5, 5, 6, 2 };
        int n = arr.Length;
        printTwoElements(arr, n);
    }
}
// This code is contributed by Sam007

PHP

 0)
            $arr[abs($arr[$i]) - 1] = 
            - $arr[abs($arr[$i]) - 1];
        else
            echo ( abs($arr[$i]));
    }
  
    echo "\nand the missing element is ";
    for($i = 0; $i < $size; $i++)
    {
        if($arr[$i] > 0)
            echo($i + 1);
    }
}
      
    // Driver Code
    $arr = array(7, 3, 4, 5, 5, 6, 2);
    $n = count($arr);
    printTwoElements($arr, $n);
  
// This code is contributed by anuj_67.
?>

Javascript

// Program to Find the repeating
// and missing elements
   
    function  printTwoElements(arr,size)
   { 
    var i;
        document.write(" The repeating element is ");
             
    for (i = 0; i < size; i++) 
    {
        var abs_value = Math.abs(arr[i]);
        if (arr[abs_value - 1] > 0)
            arr[abs_value - 1] = -arr[abs_value - 1];
        else
            document.write( abs_value);
    }
               
    document.write("<br> and the missing element is ");
    for (i = 0; i < size; i++) 
    {
        if (arr[i] > 0)
            document.write (i + 1);
    }
}
   
/* Driver code */
    arr = new Array ( 7, 3, 4, 5, 5, 6, 2 );
    n = arr.length;
    printTwoElements(arr, n);
      
    // This code is contributed by simranarora5sos

---

---

输出：

The repeating element is 5
and the missing element is 1

时间复杂度： O(n)
感谢Manish Mishra提出了此方法。

方法4(使两个方程式)
方法：

• 令x为缺失元素，y为重复元素。
• 使用公式S = n(n + 1)/ 2 – x + y获得所有数字的总和
• 使用公式P = 1 * 2 * 3 * … * n * y / x获得所有数字的乘积
• 上面的两个步骤为我们提供了两个方程，我们可以求解方程并获得x和y的值。

时间复杂度： O(n)
感谢消失的建议这个解决方案。

注意：当我们计算所有数组元素的乘积和总和时，此方法可能导致算术溢出。

方法5(使用XOR)

方法：

• 令x和y为所需的输出元素。
• 计算所有数组元素的XOR。

xor1 = arr[0]^arr[1]^arr[2]…..arr[n-1]

• 将结果与1到n之间的所有数字进行XOR运算

xor1 = xor1^1^2^…..^n

• 在结果xor1中，除x和y之外，所有元素都将彼此作废。在xor1中设置的所有位都将在x或y中设置。因此，如果采用xor1的任何置位(我们已在代码中选择了最右边的置位)，并将数组的元素分为两组-一组具有相同位的元素，而另一组未设置相同的位。这样，我们将在一组中获得x，在另一组中获得y。现在，如果我们对第一个集合中的所有元素进行XOR，我们将得到x，而在其他集合中进行同样的操作，我们将得到y。

下面是上述方法的实现：

C++

// C++ program to Find the repeating
// and missing elements
  
#include 
using namespace std;
  
/* The output of this function is stored at
*x and *y */
void getTwoElements(int arr[], int n,
                    int* x, int* y)
{
    /* Will hold xor of all elements 
    and numbers from 1 to n */
    int xor1;
  
    /* Will have only single set bit of xor1 */
    int set_bit_no;
  
    int i;
    *x = 0;
    *y = 0;
  
    xor1 = arr[0];
  
    /* Get the xor of all array elements */
    for (i = 1; i < n; i++)
        xor1 = xor1 ^ arr[i];
  
    /* XOR the previous result with numbers 
    from 1 to n*/
    for (i = 1; i <= n; i++)
        xor1 = xor1 ^ i;
  
    /* Get the rightmost set bit in set_bit_no */
    set_bit_no = xor1 & ~(xor1 - 1);
  
    /* Now divide elements into two 
    sets by comparing a rightmost set
    bit of xor1 with the bit at the same 
    position in each element. Also, 
    get XORs of two sets. The two
    XORs are the output elements. 
    The following two for loops 
    serve the purpose */
    for (i = 0; i < n; i++) {
        if (arr[i] & set_bit_no)
            /* arr[i] belongs to first set */
            *x = *x ^ arr[i];
  
        else
            /* arr[i] belongs to second set*/
            *y = *y ^ arr[i];
    }
    for (i = 1; i <= n; i++) {
        if (i & set_bit_no)
            /* i belongs to first set */
            *x = *x ^ i;
  
        else
            /* i belongs to second set*/
            *y = *y ^ i;
    }
  
    /* *x and *y hold the desired
        output elements */
}
  
/* Driver code */
int main()
{
    int arr[] = { 1, 3, 4, 5, 5, 6, 2 };
    int* x = (int*)malloc(sizeof(int));
    int* y = (int*)malloc(sizeof(int));
    int n = sizeof(arr) / sizeof(arr[0]);
  
    getTwoElements(arr, n, x, y);
    cout << " The missing element is " << *x << " and the repeating"
         << " number is " << *y;
    getchar();
}
  
// This code is contributed by Code_Mech

C

// C program to Find the repeating
// and missing elements
  
#include 
#include 
  
/* The output of this function is stored at
   *x and *y */
void getTwoElements(int arr[], int n, int* x, int* y)
{
    /* Will hold xor of all elements and numbers 
       from 1 to n */
    int xor1;
  
    /* Will have only single set bit of xor1 */
    int set_bit_no;
  
    int i;
    *x = 0;
    *y = 0;
  
    xor1 = arr[0];
  
    /* Get the xor of all array elements */
    for (i = 1; i < n; i++)
        xor1 = xor1 ^ arr[i];
  
    /* XOR the previous result with numbers 
       from 1 to n*/
    for (i = 1; i <= n; i++)
        xor1 = xor1 ^ i;
  
    /* Get the rightmost set bit in set_bit_no */
    set_bit_no = xor1 & ~(xor1 - 1);
  
    /* Now divide elements in two sets by comparing 
    rightmost set bit of xor1 with bit at same 
    position in each element. Also, get XORs of two 
    sets. The two XORs are the output elements. The
    following two for loops serve the purpose */
    for (i = 0; i < n; i++) {
        if (arr[i] & set_bit_no)
            /* arr[i] belongs to first set */
            *x = *x ^ arr[i];
  
        else
            /* arr[i] belongs to second set*/
            *y = *y ^ arr[i];
    }
    for (i = 1; i <= n; i++) {
        if (i & set_bit_no)
            /* i belongs to first set */
            *x = *x ^ i;
  
        else
            /* i belongs to second set*/
            *y = *y ^ i;
    }
  
    /* *x and *y hold the desired output elements */
}
  
/* Driver program to test above function */
int main()
{
    int arr[] = { 1, 3, 4, 5, 5, 6, 2 };
    int* x = (int*)malloc(sizeof(int));
    int* y = (int*)malloc(sizeof(int));
    int n = sizeof(arr) / sizeof(arr[0]);
  
    getTwoElements(arr, n, x, y);
    printf(" The missing element is %d"
           " and the repeating number"
           " is %d",
           *x, *y);
    getchar();
}

Java

// Java program to Find the repeating
// and missing elements
  
import java.io.*;
  
class GFG {
    static int x, y;
  
    static void getTwoElements(int arr[], int n)
    {
        /* Will hold xor of all elements
       and numbers from 1 to n  */
        int xor1;
  
        /* Will have only single set bit of xor1 */
        int set_bit_no;
  
        int i;
        x = 0;
        y = 0;
  
        xor1 = arr[0];
  
        /* Get the xor of all array elements  */
        for (i = 1; i < n; i++)
            xor1 = xor1 ^ arr[i];
  
        /* XOR the previous result with numbers from 
       1 to n*/
        for (i = 1; i <= n; i++)
            xor1 = xor1 ^ i;
  
        /* Get the rightmost set bit in set_bit_no */
        set_bit_no = xor1 & ~(xor1 - 1);
  
        /* Now divide elements into two sets by comparing
    rightmost set bit of xor1 with the bit at the same 
    position in each element. Also, get XORs of two
    sets. The two XORs are the output elements. The 
    following two for loops serve the purpose */
        for (i = 0; i < n; i++) {
            if ((arr[i] & set_bit_no) != 0)
                /* arr[i] belongs to first set */
                x = x ^ arr[i];
  
            else
                /* arr[i] belongs to second set*/
                y = y ^ arr[i];
        }
        for (i = 1; i <= n; i++) {
            if ((i & set_bit_no) != 0)
                /* i belongs to first set */
                x = x ^ i;
  
            else
                /* i belongs to second set*/
                y = y ^ i;
        }
  
        /* *x and *y hold the desired output elements */
    }
    /* Driver program to test above function */
    public static void main(String[] args)
    {
        int arr[] = { 1, 3, 4, 5, 1, 6, 2 };
  
        int n = arr.length;
        getTwoElements(arr, n);
        System.out.println(" The missing element is  "
                           + x + "and the "
                           + "repeating number is "
                           + y);
    }
}
  
// This code is contributed by Gitanjali.

Python3

# Python3 program to find the repeating 
# and missing elements 
  
# The output of this function is stored 
# at x and y 
def getTwoElements(arr, n):
      
    global x, y
    x = 0
    y = 0
      
    # Will hold xor of all elements 
    # and numbers from 1 to n 
    xor1 = arr[0]
      
    # Get the xor of all array elements
    for i in range(1, n):
        xor1 = xor1 ^ arr[i]
          
    # XOR the previous result with numbers 
    # from 1 to n
    for i in range(1, n + 1):
        xor1 = xor1 ^ i
      
    # Will have only single set bit of xor1
    set_bit_no = xor1 & ~(xor1 - 1)
      
    # Now divide elements into two 
    # sets by comparing a rightmost set 
    # bit of xor1 with the bit at the same 
    # position in each element. Also, 
    # get XORs of two sets. The two 
    # XORs are the output elements. 
    # The following two for loops 
    # serve the purpose
    for i in range(n):
        if (arr[i] & set_bit_no) != 0:
              
            # arr[i] belongs to first set
            x = x ^ arr[i]
        else:
              
            # arr[i] belongs to second set
            y = y ^ arr[i]
              
    for i in range(1, n + 1):
        if (i & set_bit_no) != 0:
              
            # i belongs to first set
            x = x ^ i
        else:
              
            # i belongs to second set
            y = y ^ i 
          
    # x and y hold the desired 
    # output elements 
      
# Driver code
arr = [ 1, 3, 4, 5, 5, 6, 2 ]
n = len(arr)
      
getTwoElements(arr, n)
  
print("The missing element is", x,
      "and the repeating number is", y)
      
# This code is contributed by stutipathak31jan

C＃

// C# program to Find the repeating
// and missing elements
  
using System;
  
class GFG {
    static int x, y;
  
    static void getTwoElements(int[] arr, int n)
    {
        /* Will hold xor of all elements
        and numbers from 1 to n */
        int xor1;
  
        /* Will have only single set bit of xor1 */
        int set_bit_no;
  
        int i;
        x = 0;
        y = 0;
  
        xor1 = arr[0];
  
        /* Get the xor of all array elements */
        for (i = 1; i < n; i++)
            xor1 = xor1 ^ arr[i];
  
        /* XOR the previous result with numbers from 
        1 to n*/
        for (i = 1; i <= n; i++)
            xor1 = xor1 ^ i;
  
        /* Get the rightmost set bit in set_bit_no */
        set_bit_no = xor1 & ~(xor1 - 1);
  
        /* Now divide elements in two sets by comparing
        rightmost set bit of xor1 with bit at same 
        position in each element. Also, get XORs of two
        sets. The two XORs are the output elements.The 
        following two for loops serve the purpose */
        for (i = 0; i < n; i++) {
            if ((arr[i] & set_bit_no) != 0)
  
                /* arr[i] belongs to first set */
                x = x ^ arr[i];
  
            else
  
                /* arr[i] belongs to second set*/
                y = y ^ arr[i];
        }
        for (i = 1; i <= n; i++) {
            if ((i & set_bit_no) != 0)
  
                /* i belongs to first set */
                x = x ^ i;
  
            else
  
                /* i belongs to second set*/
                y = y ^ i;
        }
  
        /* *x and *y hold the desired output elements */
    }
  
    // Driver program
    public static void Main()
    {
        int[] arr = { 1, 3, 4, 5, 1, 6, 2 };
  
        int n = arr.Length;
        getTwoElements(arr, n);
        Console.Write(" The missing element is "
                      + x + "and the "
                      + "repeating number is "
                      + y);
    }
}
  
// This code is contributed by Sam007

输出：

The missing element is 7 and the repeating number is 5

时间复杂度： O(n)
此方法不会引起溢出，但不会告诉您哪一个发生了两次而哪一个丢失了。我们可以再增加一个步骤，以检查缺少的一个和重复的一个。这很容易在O(n)时间内完成。

方法6(使用地图)
方法：
此方法涉及在Map的帮助下创建Hashtable。在这种情况下，元素被映射到其自然索引。在此过程中，如果一个元素被映射两次，那么它就是重复元素。而且，如果不存在元素的映射，则它是缺少的元素。

下面是上述方法的实现：

C++

// C++ program to find the repeating
// and missing elements using Maps 
#include 
#include 
using namespace std;
  
int main()
{
    int arr[] = { 4, 3, 6, 2, 1, 1 };
    int n = 6;
      
    unordered_map numberMap;
      
    for(int i : arr) 
    {
        if (numberMap.find(i) == 
            numberMap.end())
        {
            numberMap[i] = true;
        }
        else
        {
            cout << "Repeating = " << i;
        }
    }
    cout << endl;
      
    for(int i = 1; i <= n; i++)
    {
        if (numberMap.find(i) == 
            numberMap.end()) 
        {
            cout << "Missing = " << i;
        }
    }
    return 0;
}
  
// This code is contributed by RohitOberoi

Java

// Java program to find the
// repeating and missing elements
// using Maps
  
import java.util.*;
  
public class Test1 {
  
    public static void main(String[] args)
    {
  
        int[] arr = { 4, 3, 6, 2, 1, 1 };
  
        Map numberMap
            = new HashMap<>();
  
        int max = arr.length;
  
        for (Integer i : arr) {
  
            if (numberMap.get(i) == null) {
                numberMap.put(i, true);
            }
            else {
                System.out.println("Repeating = " + i);
            }
        }
        for (int i = 1; i <= max; i++) {
            if (numberMap.get(i) == null) {
                System.out.println("Missing = " + i);
            }
        }
    }
}

Python3

# Python3 program to find the 
# repeating and missing elements 
# using Maps
def main():
      
    arr = [ 4, 3, 6, 2, 1, 1 ]
      
    numberMap = {}
      
    max = len(arr)
    for i in arr:
        if not i in numberMap:
            numberMap[i] = True
              
        else:
            print("Repeating =", i)
      
    for i in range(1, max + 1):
        if not i in numberMap:
            print("Missing =", i)
main()
  
# This code is contributed by stutipathak31jan

C＃

// C# program to find the
// repeating and missing elements
// using Maps
using System;
using System.Collections.Generic;
  
class GFG
{
    public static void Main(String[] args)
    {
        int[] arr = { 4, 3, 6, 2, 1, 1 };
  
        Dictionary numberMap =
                   new Dictionary();
  
        int max = arr.Length;
  
        foreach (int i in arr) 
        {
            if (!numberMap.ContainsKey(i)) 
            {
                numberMap.Add(i, true);
            }
            else 
            {
                Console.WriteLine("Repeating = " + i);
            }
        }
        for (int i = 1; i <= max; i++) 
        {
            if (!numberMap.ContainsKey(i)) 
            {
                Console.WriteLine("Missing = " + i);
            }
        }
    }
}
  
// This code is contributed by PrinciRaj1992

输出：

Repeating = 1
Missing = 5

方法7(使用和和平方和制作两个方程式)
方法：

• 令x为缺失元素，y为重复元素。
• 令N为数组的大小。
• 使用公式S = N(N + 1)/ 2获得所有数字的总和
• 使用公式Sum_Sq = N(N + 1)(2N + 1)/ 6获得所有数字的平方和
• 从i = 1….N循环遍历
• S-= A [i]
• Sum_Sq-=(A [i] * A [i])
• 它将给出两个方程
  xy = S –(1)
  x ^ 2 – y ^ 2 = Sum_sq
  x + y =(Sum_sq / S)–(2)

时间复杂度： O(n)

C++

#include  
  
using namespace std;
  
vectorrepeatedNumber(const vector &A) {
    long long int len = A.size();
    long long int Sum_N = (len * (len+1) ) /2, Sum_NSq = (len * (len +1) *(2*len +1) )/6;
    long long int missingNumber=0, repeating=0;
      
    for(int i=0;i ans;
    ans.push_back(repeating);
    ans.push_back(missingNumber);
    return ans;
      
}
  
  
int main(void){
        std::vector v = {4, 3, 6, 2, 1, 6,7};
    vector res = repeatedNumber(v);
    for(int x: res){
        cout<< x<<"  ";
    }
    cout<

Java

import java.util.*;
  
class GFG 
{
    static Vector repeatedNumber(int[] A) 
    {
        int len = A.length;
        int Sum_N = (len * (len + 1)) / 2;
        int Sum_NSq = (len * (len + 1) * 
                         (2 * len + 1)) / 6;
        int missingNumber = 0, repeating = 0;
  
        for (int i = 0; i < A.length; i++) 
        {
            Sum_N -= A[i];
            Sum_NSq -= A[i] * A[i];
        }
  
        missingNumber = (Sum_N + Sum_NSq / 
                                 Sum_N) / 2;
        repeating = missingNumber - Sum_N;
        Vector ans = new Vector<>();
        ans.add(repeating);
        ans.add(missingNumber);
        return ans;
    }
  
    // Driver Code
    public static void main(String[] args) 
    {
        int[] v = { 4, 3, 6, 2, 1, 6, 7 };
        Vector res = repeatedNumber(v);
        for (int x : res) 
        {
            System.out.print(x + " ");
        }
    }
}
  
// This code is contributed by Rajput-Ji

Python3

def repeatedNumber(A):
      
    length = len(A)
    Sum_N = (length * (length + 1)) // 2
    Sum_NSq = ((length * (length + 1) * 
                     (2 * length + 1)) // 6)
      
    missingNumber, repeating = 0, 0
      
    for i in range(len(A)):
        Sum_N -= A[i]
        Sum_NSq -= A[i] * A[i]
          
    missingNumber = (Sum_N + Sum_NSq // 
                             Sum_N) // 2
    repeating = missingNumber - Sum_N
      
    ans = []
    ans.append(repeating)
    ans.append(missingNumber)
      
    return ans
  
# Driver code
v = [ 4, 3, 6, 2, 1, 6, 7 ]
res = repeatedNumber(v)
  
for i in res:
    print(i, end = " ")
  
# This code is contributed by stutipathak31jan 

C＃

using System;
using System.Collections.Generic;
  
class GFG 
{
    static List repeatedNumber(int[] A) 
    {
        int len = A.Length;
        int Sum_N = (len * (len + 1)) / 2;
        int Sum_NSq = (len * (len + 1) * 
                        (2 * len + 1)) / 6;
        int missingNumber = 0, repeating = 0;
  
        for (int i = 0; i < A.Length; i++) 
        {
            Sum_N -= A[i];
            Sum_NSq -= A[i] * A[i];
        }
  
        missingNumber = (Sum_N + Sum_NSq / 
                                 Sum_N) / 2;
        repeating = missingNumber - Sum_N;
        List ans = new List();
        ans.Add(repeating);
        ans.Add(missingNumber);
        return ans;
    }
  
    // Driver Code
    public static void Main(String[] args) 
    {
        int[] v = { 4, 3, 6, 2, 1, 6, 7 };
        List res = repeatedNumber(v);
        foreach (int x in res) 
        {
            Console.Write(x + " ");
        }
    }
}
  
// This code is contributed by PrinciRaj1992

输出：

6 5