查找给定数组中的两个重复元素

您将获得一个包含 n+2 个元素的数组。数组的所有元素都在 1 到 n 的范围内。并且所有元素都出现一次，除了两个出现两次的数字。找到两个重复的数字。

例子：

Input: 
arr = [4, 2, 4, 5, 2, 3, 1] 
n = 5
Output:
4 2
Explanation:
The above array has n + 2 = 7 elements with all elements occurring once except 2 and 4 which occur twice. So the output should be 4 2.

方法一（基本）

使用两个循环。在外循环中，一个一个地挑选元素，并计算在内循环中被挑选的元素的出现次数。
此方法不使用问题中提供的其他有用数据，例如数字范围在 1 到 n 之间，并且只有两个重复元素。

C++

// C++ program to Find the two repeating
// elements in a given array
#include
using namespace std;
 
void printTwoRepeatNumber(int arr[], int size)
{
    int i, j, display=0;
    int visited[size];
    for(i = 0; i < size; i++)
    {
    if (visited[i] == 1)
    {
        continue;
    }
    int count = 0;
        for(j = i + 1; j < size; j++)
        {
        if(arr[i] == arr[j])
        {
            visited[j] = 1;
            ++count;
            break;
        }
        }
        if ( (count > 0) && (display < 2)){
            ++display;
        cout<<"repeating element = "<< arr[i]<

---

C

#include
#include
void printRepeating(int arr[], int size)
{
  int i, j;
  printf(" Repeating elements are ");
  for(i = 0; i < size-1; i++)
    for(j = i+1; j < size; j++)
      if(arr[i] == arr[j])
        printf(" %d ", arr[i]);
}    
 
int main()
{
  int arr[] = {4, 2, 4, 5, 2, 3, 1};
  int arr_size = sizeof(arr)/sizeof(arr[0]); 
  printRepeating(arr, arr_size);
  getchar();
  return 0;
}

Java

class RepeatElement
{
    void printRepeating(int arr[], int size)
    {
        int i, j;
        System.out.println("Repeated Elements are :");
        for (i = 0; i < size-1; i++)
        {
            for (j = i + 1; j < size; j++)
            {
                if (arr[i] == arr[j])
                    System.out.print(arr[i] + " ");
            }
        }
    }
 
    public static void main(String[] args)
    {
        RepeatElement repeat = new RepeatElement();
        int arr[] = {4, 2, 4, 5, 2, 3, 1};
        int arr_size = arr.length;
        repeat.printRepeating(arr, arr_size);
    }
}

Python3

# Python3 program to Find the two
# repeating elements in a given array
 
def printRepeating(arr, size):
 
    print("Repeating elements are ",
                         end = '')
    for i in range (0, size-1):
        for j in range (i + 1, size):
            if arr[i] == arr[j]:
                print(arr[i], end = ' ')
     
# Driver code
arr = [4, 2, 4, 5, 2, 3, 1]
arr_size = len(arr)
printRepeating(arr, arr_size)
 
# This code is contributed by Smitha Dinesh Semwal

C#

using System;
 
class GFG
{
         
    static void printRepeating(int []arr, int size)
    {
        int i, j;
         
        Console.Write("Repeated Elements are :");
        for (i = 0; i < size-1; i++)
        {
            for (j = i + 1; j < size; j++)
            {
                if (arr[i] == arr[j])
                    Console.Write(arr[i] + " ");
            }
        }
    }
    // driver code
    public static void Main()
    {
        int []arr = {4, 2, 4, 5, 2, 3, 1};
        int arr_size = arr.Length;
         
        printRepeating(arr, arr_size);
    }
}
 
// This code is contributed by Sam007

Javascript

    function printRepeating(arr , size)
    {
        var i, j;
        document.write("Repeated Elements are :");
        for (i = 0; i < size-1; i++)
        {
            for (j = i + 1; j < size; j++)
            {
                if (arr[i] == arr[j])
                    document.write(arr[i] + " ");
            }
        }
    }
 
var arr = [4, 2, 4, 5, 2, 3, 1];
var arr_size = arr.length;
printRepeating(arr, arr_size);
 
// This code is contributed by 29AjayKumar

---

---

---

输出

Repeating elements are 4 2 

时间复杂度： O(n*n)
辅助空间： O(1)

方法 2（使用 Count 数组）
遍历数组一次。遍历时，使用大小为 n 的临时数组 count[] 跟踪数组中所有元素的计数，当您看到一个已设置计数的元素时，将其打印为副本。
该方法使用问题中给出的范围来限制count[]的大小，但不使用只有两个重复元素的数据。

C++

// C++ implementation of above approach
#include 
using namespace std;
 
void printRepeating(int arr[], int size)
{
    int *count = new int[sizeof(int)*(size - 2)];
    int i;
         
    cout << " Repeating elements are ";
    for(i = 0; i < size; i++)
    {
        if(count[arr[i]] == 1)
            cout << arr[i] << " ";
        else
            count[arr[i]]++;
    }
}
 
// Driver code
int main()
{
    int arr[] = {4, 2, 4, 5, 2, 3, 1};
    int arr_size = sizeof(arr)/sizeof(arr[0]);
    printRepeating(arr, arr_size);
    return 0;
}
 
// This is code is contributed by rathbhupendra

C

#include
#include
 
void printRepeating(int arr[], int size)
{
  int *count = (int *)calloc(sizeof(int), (size - 2));
  int i;
   
  printf(" Repeating elements are ");
  for(i = 0; i < size; i++)
  { 
    if(count[arr[i]] == 1)
      printf(" %d ", arr[i]);
    else
     count[arr[i]]++;
  }   
}    
 
int main()
{
  int arr[] = {4, 2, 4, 5, 2, 3, 1};
  int arr_size = sizeof(arr)/sizeof(arr[0]); 
  printRepeating(arr, arr_size);
  getchar();
  return 0;
}

Java

class RepeatElement
{
    void printRepeating(int arr[], int size)
    {
        int count[] = new int[size];
        int i;
 
        System.out.println("Repeated elements are : ");
        for (i = 0; i < size; i++)
        {
            if (count[arr[i]] == 1)
                System.out.print(arr[i] + " ");
            else
                count[arr[i]]++;
        }
    }
 
    public static void main(String[] args)
    {
        RepeatElement repeat = new RepeatElement();
        int arr[] = {4, 2, 4, 5, 2, 3, 1};
        int arr_size = arr.length;
        repeat.printRepeating(arr, arr_size);
    }
}

Python3

# Python3 code for Find the two repeating
# elements in a given array
 
 
def printRepeating(arr,size) :
    count = [0] * size
    print(" Repeating elements are ",end = "")
    for i in range(0, size) :
        if(count[arr[i]] == 1) :
            print(arr[i], end = " ")
        else :
            count[arr[i]] = count[arr[i]] + 1
 
 
# Driver code
arr = [4, 2, 4, 5, 2, 3, 1]
arr_size = len(arr)
printRepeating(arr, arr_size)
 
 
         
# This code is contributed by Nikita Tiwari.

C＃

// C# program to Find the two
// repeating elements in a given array
using System;
 
class GFG
{
         
    static void printRepeating(int []arr,
                                    int size)
    {
        int []count = new int[size];
        int i;
 
        Console.Write("Repeated elements are: ");
        for (i = 0; i < size; i++)
        {
            if (count[arr[i]] == 1)
                Console.Write(arr[i] + " ");
            else
                count[arr[i]]++;
        }
    }
 
    // driver code
    public static void Main()
    {
        int []arr = {4, 2, 4, 5, 2, 3, 1};
        int arr_size = arr.Length;
         
        printRepeating(arr, arr_size);
    }
}
 
//This code is contributed by Sam007

Javascript

 
    // Javascript program to Find the two
    // repeating elements in a given array
       
    function printRepeating(arr, size)
    {
        let count = new Array(size);
        count.fill(0);
        let i;
  
        document.write("Repeating elements are ");
        for (i = 0; i < size; i++)
        {
            if (count[arr[i]] == 1)
                document.write(arr[i] + " ");
            else
                count[arr[i]]++;
        }
    }
     
    let arr = [4, 2, 4, 5, 2, 3, 1];
    let arr_size = arr.length;
 
    printRepeating(arr, arr_size);
     

输出

Repeating elements are 4 2 

时间复杂度： O(n)
辅助空间： O(n)

方法3（制作两个方程）

让重复的数字是 X 和 Y。我们为 X 和 Y 制作两个方程，剩下的简单任务是求解这两个方程。
我们知道从 1 到 n 的整数之和是 n(n+1)/2，乘积是 n!。当这个和从 n(n+1)/2 中减去时，我们计算输入数组的和，我们得到 X + Y，因为 X 和 Y 是集合 [1..n] 中缺少的两个数字。同样计算输入数组的乘积，当这个乘积除以 n! 时，我们得到 X*Y。给定 X 和 Y 的和和乘积，我们可以很容易地找出 X 和 Y。
设数组中所有数字的总和为 S，乘积为 P
X + Y = S – n(n+1)/2
XY = P/n！
使用上述两个等式，我们可以找出 X 和 Y。对于数组 = 4 2 4 5 2 3 1，我们得到 S = 21 和 P 为 960。
X + Y = 21 – 15 = 6
XY = 960/5！ = 8
X – Y = sqrt((X+Y)^2 – 4*XY) = sqrt(4) = 2
使用以下两个方程，我们很容易得到 X = (6 + 2)/2 和 Y = (6-2)/2
X + Y = 6
X – Y = 2
感谢 geek4u 提出这种方法。正如初学者所指出的，这种方法可能存在加法和乘法溢出问题。

方法 3 和 4 使用问题中给出的所有有用信息：

C++

#include 
using namespace std;
 
/* function to get factorial of n */
int fact(int n);
 
void printRepeating(int arr[], int size)
{
    int S = 0; /* S is for sum of elements in arr[] */
    int P = 1; /* P is for product of elements in arr[] */
    int x, y; /* x and y are two repeating elements */
    int D; /* D is for difference of x and y, i.e., x-y*/
    int n = size - 2, i;
     
    /* Calculate Sum and Product of all elements in arr[] */
    for(i = 0; i < size; i++)
    {
        S = S + arr[i];
        P = P*arr[i];
    }    
         
    S = S - n*(n+1)/2; /* S is x + y now */
    P = P/fact(n);     /* P is x*y now */
         
    D = sqrt(S*S - 4*P); /* D is x - y now */
         
    x = (D + S)/2;
    y = (S - D)/2;
         
    cout<<"The two Repeating elements are "<

C

#include
#include
#include
 
/* function to get factorial of n */
int fact(int n);
 
void printRepeating(int arr[], int size)
{
  int S = 0;  /* S is for sum of elements in arr[] */
  int P = 1;  /* P is for product of elements in arr[] */ 
  int x,  y;   /* x and y are two repeating elements */
  int D;      /* D is for difference of x and y, i.e., x-y*/
  int n = size - 2,  i;
 
  /* Calculate Sum and Product of all elements in arr[] */
  for(i = 0; i < size; i++)
  {
    S = S + arr[i];
    P = P*arr[i];
  }       
   
  S = S - n*(n+1)/2;  /* S is x + y now */
  P = P/fact(n);         /* P is x*y now */
   
  D = sqrt(S*S - 4*P); /* D is x - y now */
   
  x = (D + S)/2;
  y = (S - D)/2;
   
  printf("The two Repeating elements are %d & %d", x, y);
}    
 
int fact(int n)
{
   return (n == 0)? 1 : n*fact(n-1);
}   
 
int main()
{
  int arr[] = {4, 2, 4, 5, 2, 3, 1};
  int arr_size = sizeof(arr)/sizeof(arr[0]); 
  printRepeating(arr, arr_size);
  getchar();
  return 0;
}

Java

class RepeatElement
{
    void printRepeating(int arr[], int size)
    {
        /* S is for sum of elements in arr[] */
        int S = 0;
         
        /* P is for product of elements in arr[] */
        int P = 1;
         
        /* x and y are two repeating elements */
        int x, y;
         
        /* D is for difference of x and y, i.e., x-y*/
        int D;
         
        int n = size - 2, i;
 
        /* Calculate Sum and Product of all elements in arr[] */
        for (i = 0; i < size; i++)
        {
            S = S + arr[i];
            P = P * arr[i];
        }
 
        /* S is x + y now */
        S = S - n * (n + 1) / 2;
         
        /* P is x*y now */
        P = P / fact(n);
        
        /* D is x - y now */
        D = (int) Math.sqrt(S * S - 4 * P);
         
 
        x = (D + S) / 2;
        y = (S - D) / 2;
 
        System.out.println("The two repeating elements are :");
        System.out.print(x + " " + y);
    }
 
    int fact(int n)
    {
        return (n == 0) ? 1 : n * fact(n - 1);
    }
 
    public static void main(String[] args) {
        RepeatElement repeat = new RepeatElement();
        int arr[] = {4, 2, 4, 5, 2, 3, 1};
        int arr_size = arr.length;
        repeat.printRepeating(arr, arr_size);
    }
}
 
// This code has been contributed by Mayank Jaiswal

Python3

# Python3 code for Find the two repeating
# elements in a given array
import math
 
def printRepeating(arr, size) :
     
    # S is for sum of elements in arr[]
    S = 0;
     
    # P is for product of elements in arr[]
    P = 1;
     
    n = size - 2
 
    # Calculate Sum and Product
    # of all elements in arr[]
    for i in range(0, size) :
        S = S + arr[i]
        P = P * arr[i]
     
    # S is x + y now
    S = S - n * (n + 1) // 2
     
     # P is x*y now
    P = P // fact(n)   
     
    # D is x - y now
    D = math.sqrt(S * S - 4 * P)
     
    x = (D + S) // 2
    y = (S - D) // 2
     
    print("The two Repeating elements are ",
          (int)(x)," & " ,(int)(y))
     
 
def fact(n) :
    if(n == 0) :
        return 1
    else :
        return(n * fact(n - 1))
 
# Driver code
arr = [4, 2, 4, 5, 2, 3, 1]
arr_size = len(arr)
printRepeating(arr, arr_size)
 
         
# This code is contributed by Nikita Tiwari.

C＃

using System;
 
class GFG
{
         
    static void printRepeating(int []arr, int size)
    {
        /* S is for sum of elements in arr[] */
        int S = 0;
         
        /* P is for product of elements in arr[] */
        int P = 1;
         
        /* x and y are two repeating elements */
        int x, y;
         
        /* D is for difference of x and y, i.e., x-y*/
        int D;
         
        int n = size - 2, i;
 
        /* Calculate Sum and Product
         of all elements in arr[] */
        for (i = 0; i < size; i++)
        {
            S = S + arr[i];
            P = P * arr[i];
        }
 
        /* S is x + y now */
        S = S - n * (n + 1) / 2;
         
        /* P is x*y now */
        P = P / fact(n);
         
        /* D is x - y now */
        D = (int) Math.Sqrt(S * S - 4 * P);
         
 
        x = (D + S) / 2;
        y = (S - D) / 2;
 
        Console.WriteLine("The two" +
                " repeating elements are :");
        Console.Write(x + " " + y);
    }
 
    static int fact(int n)
    {
        return (n == 0) ? 1 : n * fact(n - 1);
    }
 
    // driver code
    public static void Main() {
        int []arr = {4, 2, 4, 5, 2, 3, 1};
        int arr_size = arr.Length;
         
        printRepeating(arr, arr_size);
    }
}
// This code is contributed by Sam007

Javascript

 
    function printRepeating(arr , size)
    {
        /* S is for sum of elements in arr */
        var S = 0;
         
        /* P is for product of elements in arr */
        var P = 1;
         
        /* x and y are two repeating elements */
        var x, y;
         
        /* D is for difference of x and y, i.e., x-y*/
        var D;
         
        var n = size - 2, i;
 
        /* Calculate Sum and Product of all elements in arr */
        for (i = 0; i < size; i++)
        {
            S = S + arr[i];
            P = P * arr[i];
        }
 
        /* S is x + y now */
        S = S - n * parseInt((n + 1) / 2);
         
        /* P is x*y now */
        P = parseInt(P / fact(n));
        
        /* D is x - y now */
        D = parseInt( Math.sqrt(S * S - 4 * P));
         
 
        x = parseInt((D + S) / 2);
        y = parseInt((S - D) / 2);
 
        document.write("The two repeating elements are : ");
        document.write(x + " & " + y);
    }
 
    function fact(n)
    {    var ans = false;
        if(n == 0)
            return 1;
        else
            return(n * fact(n - 1));
    }
 
     
    var arr = [4, 2, 4, 5, 2, 3, 1];
    var arr_size = arr.length;
    printRepeating(arr, arr_size);
 
// This code is contributed by 29AjayKumar
 

输出

The two Repeating elements are 4 & 2

时间复杂度： O(n)
辅助空间： O(1)

方法 4（使用 XOR）

感谢新手提出这种方法。
这里使用的方法类似于本文的方法 2。
令重复数为 X 和 Y，如果我们对数组中的所有元素和从 1 到 n 的所有整数进行异或，则结果为 X xor Y。
X xor Y 的二进制表示中的 1 对应于 X 和 Y 之间的不同位。假设 X xor Y 的第 k 位为 1，我们可以对数组中的所有元素和从 1 到 n 的所有整数进行异或，其第 k 位为 1。结果将是 X 和 Y 之一。

C++

#include 
using namespace std;
 
void printRepeating(int arr[], int size)
{
    int Xor = arr[0]; /* Will hold Xor of all elements */
    int set_bit_no; /* Will have only single set bit of Xor */
    int i;
    int n = size - 2;
    int x = 0, y = 0;
         
    /* Get the Xor of all elements in arr[] and {1, 2 .. n} */
    for(i = 1; i < size; i++)
        Xor ^= arr[i];
    for(i = 1; i <= n; i++)
        Xor ^= i;
     
    /* Get the rightmost set bit in set_bit_no */
    set_bit_no = Xor & ~(Xor-1);
     
    /* Now divide elements in two sets by comparing rightmost set
    bit of Xor with bit at same position in each element. */
    for(i = 0; i < size; i++)
    {
        if(arr[i] & set_bit_no)
        x = x ^ arr[i]; /*Xor of first set in arr[] */
        else
        y = y ^ arr[i]; /*Xor of second set in arr[] */
    }
    for(i = 1; i <= n; i++)
    {
        if(i & set_bit_no)
        x = x ^ i; /*Xor of first set in arr[] and {1, 2, ...n }*/
        else
        y = y ^ i; /*Xor of second set in arr[] and {1, 2, ...n } */
    }
         
    cout<<"The two repeating elements are "<

C

void printRepeating(int arr[], int size)
{
  int xor = arr[0]; /* Will hold xor of all elements */
  int set_bit_no;  /* Will have only single set bit of xor */
  int i;
  int n = size - 2;
  int x = 0, y = 0;
   
  /* Get the xor of all elements in arr[] and {1, 2 .. n} */
  for(i = 1; i < size; i++)
    xor ^= arr[i]; 
  for(i = 1; i <= n; i++)
    xor ^= i;  
 
  /* Get the rightmost set bit in set_bit_no */
  set_bit_no = xor & ~(xor-1);
 
  /* Now divide elements in two sets by comparing rightmost set
   bit of xor with bit at same position in each element. */
  for(i = 0; i < size; i++)
  {
    if(arr[i] & set_bit_no)
      x = x ^ arr[i]; /*XOR of first set in arr[] */
    else
      y = y ^ arr[i]; /*XOR of second set in arr[] */
  }
  for(i = 1; i <= n; i++)
  {
    if(i & set_bit_no)
      x = x ^ i; /*XOR of first set in arr[] and {1, 2, ...n }*/
    else
      y = y ^ i; /*XOR of second set in arr[] and {1, 2, ...n } */
  }
   
  printf("n The two repeating elements are %d & %d ", x, y);
}    
 
 
int main()
{
  int arr[] = {4, 2, 4, 5, 2, 3, 1};
  int arr_size = sizeof(arr)/sizeof(arr[0]); 
  printRepeating(arr, arr_size);
  getchar();
  return 0;
}

Java

class RepeatElement
{
    void printRepeating(int arr[], int size)
    {
        /* Will hold xor of all elements */
        int xor = arr[0];
         
        /* Will have only single set bit of xor */
        int set_bit_no;
         
        int i;
        int n = size - 2;
        int x = 0, y = 0;
 
        /* Get the xor of all elements in arr[] and {1, 2 .. n} */
        for (i = 1; i < size; i++)
            xor ^= arr[i];
        for (i = 1; i <= n; i++)
            xor ^= i;
 
        /* Get the rightmost set bit in set_bit_no */
        set_bit_no = (xor & ~(xor - 1));
 
        /* Now divide elements in two sets by comparing rightmost set
           bit of xor with bit at same position in each element. */
        for (i = 0; i < size; i++) {
            int a = arr[i] & set_bit_no;
            if (a != 0)
                x = x ^ arr[i]; /*XOR of first set in arr[] */
            else
                y = y ^ arr[i]; /*XOR of second set in arr[] */
        }
        for (i = 1; i <= n; i++)
        {
            int a = i & set_bit_no;
            if (a != 0)
                x = x ^ i; /*XOR of first set in arr[] and {1, 2, ...n }*/
            else
                y = y ^ i; /*XOR of second set in arr[] and {1, 2, ...n } */
        }
 
        System.out.println("The two reppeated elements are :");
        System.out.println(x + " " + y);
    }
 
    /* Driver program to test the above function */
    public static void main(String[] args)
    {
        RepeatElement repeat = new RepeatElement();
        int arr[] = {4, 2, 4, 5, 2, 3, 1};
        int arr_size = arr.length;
        repeat.printRepeating(arr, arr_size);
    }
}
 
// This code has been contributed by Mayank Jaiswal

Python3

# Python3 code to Find the
# two repeating elements
# in a given array
def printRepeating(arr, size):
     
    # Will hold xor
    # of all elements
    xor = arr[0]
    n = size - 2
    x = 0
    y = 0
     
    # Get the xor of all
    # elements in arr[]
    # and 1, 2 .. n
    for i in range(1 , size):
        xor ^= arr[i]
    for i in range(1 , n + 1):
        xor ^= i
     
    # Get the rightmost set
    # bit in set_bit_no
    set_bit_no = xor & ~(xor-1)
     
    # Now divide elements in two
    # sets by comparing rightmost
    # set bit of xor with bit at
    # same position in each element.
    for i in range(0, size):
         
        if(arr[i] & set_bit_no):
             
            # XOR of first
            # set in arr[]
            x = x ^ arr[i]
        else:
             
            # XOR of second
            # set in arr[]
            y = y ^ arr[i]
             
    for i in range(1 , n + 1):
 
        if(i & set_bit_no):
             
            # XOR of first set
            # in arr[] and
            # 1, 2, ...n
            x = x ^ i
        else:
             
             # XOR of second set
             # in arr[] and
             # 1, 2, ...n
            y = y ^ i
             
    print("The two repeating",
         "elements are", y, x)
 
# Driver code   
arr = [4, 2, 4,
       5, 2, 3, 1]
arr_size = len(arr)
printRepeating(arr, arr_size)
 
# This code is contributed
# by Smitha

C＃

using System;
 
class GFG
{
    static void printRepeating(int []arr, int size)
    {
        /* Will hold xor of all elements */
        int xor = arr[0];
         
        /* Will have only single set bit of xor */
        int set_bit_no;
         
        int i;
        int n = size - 2;
        int x = 0, y = 0;
 
        /* Get the xor of all elements
         in arr[] and {1, 2 .. n} */
        for (i = 1; i < size; i++)
            xor ^= arr[i];
         
        for (i = 1; i <= n; i++)
            xor ^= i;
 
        /* Get the rightmost set bit in set_bit_no */
        set_bit_no = (xor & ~(xor - 1));
 
        /* Now divide elements in two sets by
         comparing rightmost set bit of xor with bit
         at same position in each element. */
        for (i = 0; i < size; i++) {
            int a = arr[i] & set_bit_no;
         
            if (a != 0)
                /* XOR of first set in arr[] */
                x = x ^ arr[i];
            else
                /* XOR of second set in arr[] */
                y = y ^ arr[i];
        }
        for (i = 1; i <= n; i++)
        {
            int a = i & set_bit_no;
             
            if (a != 0)
             
                /* XOR of first set in
                 arr[] and {1, 2, ...n }*/
                x = x ^ i;
            else
             
                /* XOR of second set in
                 arr[] and {1, 2, ...n } */
                y = y ^ i;
        }
 
        Console.WriteLine("The two" +
            " reppeated elements are :");
        Console.Write(x + " " + y);
    }
 
    /* Driver program to test the above function */
    public static void Main()
    {
        int []arr = {4, 2, 4, 5, 2, 3, 1};
        int arr_size = arr.Length;
         
        printRepeating(arr, arr_size);
    }
}
 
// This code is contributed by Sam007

Javascript

 
function printRepeating( arr, size)
{
    let Xor = arr[0]; /* Will hold Xor of all elements */
    let set_bit_no; /* Will have only single set bit of Xor */
    let i;
    let n = size - 2;
    let x = 0, y = 0;
         
    /* Get the Xor of all elements in arr[] and {1, 2 .. n} */
    for(i = 1; i < size; i++)
        Xor ^= arr[i];
    for(i = 1; i <= n; i++)
        Xor ^= i;
     
    /* Get the rightmost set bit in set_bit_no */
    set_bit_no = Xor & ~(Xor-1);
     
    /* Now divide elements in two sets by comparing rightmost set
    bit of Xor with bit at same position in each element. */
    for(i = 0; i < size; i++)
    {
        if(arr[i] & set_bit_no)
        x = x ^ arr[i]; /*Xor of first set in arr[] */
        else
        y = y ^ arr[i]; /*Xor of second set in arr[] */
    }
    for(i = 1; i <= n; i++)
    {
        if(i & set_bit_no)
        x = x ^ i; /*Xor of first set in arr[] and {1, 2, ...n }*/
        else
        y = y ^ i; /*Xor of second set in arr[] and {1, 2, ...n } */
    }
         
    document.write("The two repeating elements are "+y+" "+x);
}
    // driver code
    let arr = [4, 2, 4, 5, 2, 3, 1];
    let arr_size = arr.length;
    printRepeating(arr, arr_size);
     
    // This code is contributed by jana_sayantan.

输出

The two repeating elements are 4 2

方法五（使用数组元素作为索引）
感谢 Manish K. Aasawat 提出这种方法。

Traverse the array. Do following for every index i of A[].
{
check for sign of A[abs(A[i])] ;
if positive then
   make it negative by   A[abs(A[i])]=-A[abs(A[i])];
else  // i.e., A[abs(A[i])] is negative
   this   element (ith element of list) is a repetition
}

Example: A[] =  {1, 1, 2, 3, 2}
i=0; 
Check sign of A[abs(A[0])] which is A[1].  A[1] is positive, so make it negative. 
Array now becomes {1, -1, 2, 3, 2}

i=1; 
Check sign of A[abs(A[1])] which is A[1].  A[1] is negative, so A[1] is a repetition.

i=2; 
Check sign of A[abs(A[2])] which is A[2].  A[2] is  positive, so make it negative. '
Array now becomes {1, -1, -2, 3, 2}

i=3; 
Check sign of A[abs(A[3])] which is A[3].  A[3] is  positive, so make it negative. 
Array now becomes {1, -1, -2, -3, 2}

i=4; 
Check sign of A[abs(A[4])] which is A[2].  A[2] is negative, so A[4] is a repetition.

请注意，此方法会修改原始数组，如果我们不允许修改数组，则可能不是推荐的方法。

C++

#include 
using namespace std;
 
void printRepeating(int arr[], int size)
{
    int i;
     
    cout << "The repeating elements are";
         
    for(i = 0; i < size; i++)
    {
        if(arr[abs(arr[i])] > 0)
            arr[abs(arr[i])] = -arr[abs(arr[i])];
        else
            cout<<" " << abs(arr[i]) <<" ";
    }    
}
 
// Driver code
int main()
{
    int arr[] = {4, 2, 4, 5, 2, 3, 1};
    int arr_size = sizeof(arr)/sizeof(arr[0]);
    printRepeating(arr, arr_size);
    return 0;
}
 
// This code is contributed by rathbhupendra

C

#include 
#include 
 
void printRepeating(int arr[], int size)
{
  int i; 
  
  printf("\n The repeating elements are");
   
  for(i = 0; i < size; i++)
  {
    if(arr[abs(arr[i])] > 0)
      arr[abs(arr[i])] = -arr[abs(arr[i])];
    else
      printf(" %d ", abs(arr[i]));
  }        
}    
 
int main()
{
  int arr[] = {4, 2, 4, 5, 2, 3, 1};
  int arr_size = sizeof(arr)/sizeof(arr[0]);
  printRepeating(arr, arr_size);
  getchar();
  return 0;
}

Java

class RepeatElement
{
    void printRepeating(int arr[], int size)
    {
        int i; 
        System.out.println("The repeating elements are : ");
    
        for(i = 0; i < size; i++)
        {
           int abs_val = Math.abs(arr[i]);
            if(arr[abs_val] > 0)
                arr[abs_val] = -arr[abs_val];
            else
                System.out.print(abs_val + " ");
        }        
    }
 
    /* Driver program to test the above function */
    public static void main(String[] args)
    {
        RepeatElement repeat = new RepeatElement();
        int arr[] = {4, 2, 4, 5, 2, 3, 1};
        int arr_size = arr.length;
        repeat.printRepeating(arr, arr_size);
    }
}
 
// This code has been contributed by Mayank Jaiswal

Python3

# Python3 code for Find the two repeating
# elements in a given array
 
 
def printRepeating(arr, size) :
     
    print(" The repeating elements are",end=" ")
     
    for i in range(0,size) :
         
        if(arr[abs(arr[i])] > 0) :
            arr[abs(arr[i])] = (-1) * arr[abs(arr[i])]
        else :
            print(abs(arr[i]),end = " ")
 
# Driver code
arr = [4, 2, 4, 5, 2, 3, 1]
arr_size = len(arr)
printRepeating(arr, arr_size)
 
 
# This code is contributed by Nikita Tiwari.

C＃

// C# code for Find the two repeating
// elements in a given array
 using System;
 
class GFG
{
    static void printRepeating(int []arr, int size)
    {
        int i;
        Console.Write("The repeating elements are : ");
     
        for(i = 0; i < size; i++)
        {
          int abs_val = Math.Abs(arr[i]);
            if(arr[abs_val] > 0)
                arr[abs_val] = -arr[abs_val];
            else
                Console.Write(abs_val + " ");
        }        
    }
 
    /* Driver program to test the above function */
    public static void Main()
    {
        int []arr = {4, 2, 4, 5, 2, 3, 1};
        int arr_size = arr.Length;
         
        printRepeating(arr, arr_size);
    }
}
 
// This code is contributed by Sam007

PHP

 0)
    $arr[abs($arr[$i])] = -$arr[abs($arr[$i])];
    else
    echo  abs($arr[$i])," ";
}        
}    
 
 
    $arr = array (4, 2, 4, 5, 2, 3, 1);
    $arr_size = sizeof($arr);
    printRepeating($arr, $arr_size);
     
     
#This code is contributed by aj_36
?>

Javascript

 
function printRepeating(arr , size)
{
    var i; 
    document.write("The repeating elements are : ");
    
        for(i = 0; i < size; i++)
        {
            var abs_val = Math.abs(arr[i]);
            if(arr[abs_val] > 0)x
                arr[abs_val] = -arr[abs_val];
            else
                document.write(abs_val + " ");
    }        
}
 
/* Driver program to test the above function */
  
var arr = [4, 2, 4, 5, 2, 3, 1];
var arr_size = arr.length;
printRepeating(arr, arr_size);
 
// This code is contributed by 29AjayKumar
 

输出

The repeating elements are 4  2 

方法6（类似于方法5）

感谢 Vivek Kumar 提出这种方法。

关键是将第 (arr[i]th-1) 索引处的每个元素增加 N-1（因为元素仅存在 N-2），同时检查该索引处的元素是否除以 ( N-1) 给出 2。如果这是真的，那么这意味着该元素已经出现了两次，我们可以轻松地说这是我们的答案之一。当我们想要计算有限范围数组元素的频率时，这是非常有用的技术之一（请参阅这篇文章了解更多信息）。

This method works on the fact that the remainder of an element when divided by any number greater than that element will always give that same element. Similarly, as we are incrementing each element at (arr[i]th-1) index by N-1, so when this incremented element is divided by N-1 it will give number of times we have added N-1 to it, hence giving us the occurrence of that index (according to 1-based indexing).

下面给出的代码应用了这种方法：

C++

#include 
using namespace std;
 
void twoRepeated(int arr[], int N)
{
    int m = N - 1;
    for (int i = 0; i < N; i++) {
        arr[arr[i] % m - 1] += m;
        if ((arr[arr[i] % m - 1] / m) == 2)
            cout << arr[i] % m << " ";
    }
}
// Driver Code
int main()
{
    int arr[] = { 4, 2, 4, 5, 2, 3, 1 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << "The two repeating elements are ";
    twoRepeated(arr, n);
 
    return 0;
}
// This code is contributed by Kartik Singh Kushwah

Java

import java.io.*;
 
class GFG{
     
public static void twoRepeated(int arr[], int N)
{
    int m = N - 1;
    for(int i = 0; i < N; i++)
    {
        arr[arr[i] % m - 1] += m;
         
        if ((arr[arr[i] % m - 1] / m) == 2)
            System.out.printf(arr[i] % m + " ");
    }
}
 
// Driver code
public static void main(String[] args)
{
    int arr[] = { 4, 2, 4, 5, 2, 3, 1 };
    int n = 7;
     
    System.out.printf("The two repeating elements are ");
     
    twoRepeated(arr, n);
}
}
 
// This code is contributed by Potta Lokesh

Python3

# Python program for the above approach
def twoRepeated(arr, N):
    m = N - 1
    for i in range(N):
        arr[arr[i] % m - 1] += m
        if ((arr[arr[i] % m - 1] // m) == 2):
            print(arr[i] % m ,end= " ")
     
# Driver Code
arr = [4, 2, 4, 5, 2, 3, 1]
n = len(arr)
print("The two repeating elements are ", end = "")
twoRepeated(arr, n)
 
# This code is contributed by Shubham Singh

C＃

// C# program for the above approach
using System;
 
public class GFG
{
public static void twoRepeated(int[] arr, int N)
{
    int m = N - 1;
    for(int i = 0; i < N; i++)
    {
        arr[arr[i] % m - 1] += m;
         
        if ((arr[arr[i] % m - 1] / m) == 2)
            Console.Write(arr[i] % m + " ");
    }
}
 
// Driver code
public static void Main(String []args)
{
    int[] arr = { 4, 2, 4, 5, 2, 3, 1 };
    int n = 7;
     
    Console.Write("The two repeating elements are ");
     
    twoRepeated(arr, n);
}
}
 
// This code is contributed by splevel62.

Javascript

 
function twoRepeated(arr, N)
{
    let m = N - 1;
    for(let i = 0; i < N; i++)
    {
        arr[parseInt(arr[i] % m) - 1] += m;
        if (parseInt(arr[parseInt(arr[i] % m) - 1] / m) == 2)
            document.write(parseInt(arr[i] % m) + " ");
    }
}
 
// Driver Code
var arr = [ 4, 2, 4, 5, 2, 3, 1 ];
var n = 7;
document.write("The two repeating elements are ");
 
twoRepeated(arr, n);
 
// This code is contributed by Potta Lokesh
 

输出

The two repeating elements are 4 2 

方法 7
这里的重点是将数组元素一个一个地输入到无序集合中。如果集合中已经存在特定元素，则它是重复元素。

C++

// C++ program to Find the two repeating
// elements in a given array
#include 
using namespace std;
 
void printRepeating(int arr[], int size)
{
      unordered_sets;
      cout<<"The two Repeating elements are : ";
    for(int i=0;i

Java

// Java program to Find the two repeating
// elements in a given array
import java.util.*;
class GFG{
 
  static void printRepeating(int arr[], int size)
  {
    HashSets = new HashSet<>();
    System.out.print("The two Repeating elements are : ");
    for(int i = 0; i < size; i++)
    {
      if(!s.isEmpty() && s.contains(arr[i]))
        System.out.print(arr[i]+"  ");
      s.add(arr[i]);
    }
  }
 
  // Driver code
  public static void main(String[] args)
  {
    int arr[] = {4, 2, 4, 5, 2, 3, 1};
    int arr_size = arr.length;
    printRepeating(arr, arr_size);
  }
}
 
// This code is contributed by Rajput-Ji

Python3

# Python3 program to find the two repeating
# elements in a given array
def printRepeating(arr, size):
     
    s = set()
    print("The two Repeating elements are : ", end = "")
     
    for i in range(size):
        if (len(s) and arr[i] in s):
            print(arr[i], end = "  ")
             
        s.add(arr[i])
         
# Driver code
arr = [ 4, 2, 4, 5, 2, 3, 1 ]
arr_size = len(arr)
printRepeating(arr, arr_size)
 
# This code is contributed by Shubham Singh

C＃

// C# program to Find the two repeating
// elements in a given array
using System;
using System.Collections.Generic;
 
public class GFG{
     
  static void printRepeating(int[] arr, int size)
  {
    HashSets = new HashSet();
    Console.Write("The two Repeating elements are : ");
    for(int i = 0; i < size; i++)
    {
      if(s.Count != 0 && s.Contains(arr[i]))
        Console.Write(arr[i] + "  ");
      s.Add(arr[i]);
    }
  }
 
  // Driver code
  public static void Main()
  {
    int[] arr = {4, 2, 4, 5, 2, 3, 1};
    int arr_size = arr.Length;
    printRepeating(arr, arr_size);
  }
}
 
// This code is contributed by Shubham Singh

Javascript

 
function printRepeating(arr, size)
{
    const s = new Set();
    document.write("The two repeating elements are ");
    for(let i = 0; i < size; i++)
    {
      if(s.size != 0 && s.has(arr[i]))
        document.write(arr[i] + " ");
      s.add(arr[i]);
             
    }
}
 
// Driver Code
var arr = [ 4, 2, 4, 5, 2, 3, 1 ];
var arr_size = 7;
printRepeating(arr, arr_size);
 
// This code is contributed by Shubham Singh
 

输出

The two Repeating elements are : 4  2  

时间复杂度：O(n)

辅助空间：O(n)