在有限范围的数组中查找偶发的元素

给定一个数组，其中包含所有数字的奇数个出现次数，除了少数元素出现偶数次之外。在O(n)时间复杂度和O(1)额外空间中找到在数组中偶数出现的元素。
假设数组包含0到63范围内的元素。
例子 ：

Input: [9, 12, 23, 10, 12, 12, 15, 23, 14, 12, 15]
Output: 12, 23 and 15

Input: [1, 4, 7, 5, 9, 7, 3, 4, 6, 8, 3, 0, 3]
Output: 4 and 7

Input: [4, 4, 10, 10, 4, 4, 4, 4, 10, 10]
Output: 4 and 10

一种简单的方法是遍历数组并将其元素的频率存储在地图中。以后，打印甚至在地图中计数的元素。该解决方案花费O(n)时间，但需要额外的空间来存储频率。下面是一种使用按位运算运算符解决此问题的有趣方法。
此方法假定使用64位存储long long整数。这个想法是使用XOR运算符。我们知道

1 XOR 1 = 0
1 XOR 0 = 1
0 XOR 1 = 1
0 XOR 0 = 0

考虑以下输入–

1, 4, 7, 5, 9, 7, 3, 4, 6, 8, 3, 0, 3 

如果我们将数组的每个元素的值左移1，并对所有项进行XOR，我们将得到二进制整数以下的值–

1101101011

从右起第i个索引中的每个1表示元素i的奇数出现。从右起第i个索引中的每个0表示数组中元素i的偶数或不出现。
在上述二进制数的第二，第四和第七位置中存在0。但是2在我们的数组中不存在。因此，我们的答案是4和7。
以下是上述想法的实现

C++

// C++ Program to find the even occurring elements
// in given array
#include 
using namespace std;
 
// Function to find the even occurring elements
// in given array
void printRepeatingEven(int arr[], int n)
{
    long long _xor = 0L;
    long long pos;
 
    // do for each element of array
    for( int i = 0; i < n; ++i)
    {
        // left-shift 1 by value of current element
        pos = 1 << arr[i];
 
        // Toggle the bit everytime element gets repeated
        _xor ^= pos;
    }
 
    // Traverse array again and use _xor to find even
    // occurring elements
    for (int i = 0; i < n; ++i)
    {
        // left-shift 1 by value of current element
        pos = 1 << arr[i];
 
        // Each 0 in _xor represents an even occurrence
        if (!(pos & _xor))
        {
            // print the even occurring numbers
            cout << arr[i] << " ";
 
            // set bit as 1 to avoid printing duplicates
            _xor ^= pos;
        }
    }
}
 
// Driver code
int main()
{
    int arr[] = { 9, 12, 23, 10, 12, 12, 15, 23,
                 14, 12, 15 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    printRepeatingEven(arr, n);
 
    return 0;
}

Java

// Java Program to find the even occurring
// elements in given array
class GFG
{
 
// Function to find the even occurring
// elements in given array
static void printRepeatingEven(int arr[],
                               int n)
{
    long _xor = 0L;
    long pos;
 
    // do for each element of array
    for (int i = 0; i < n; ++i)
    {
        // left-shift 1 by value of
        // current element
        pos = 1 << arr[i];
 
        // Toggle the bit everytime
        // element gets repeated
        _xor ^= pos;
    }
 
    // Traverse array again and use _xor
    // to find even occurring elements
    for (int i = 0; i < n; ++i)
    {
        // left-shift 1 by value of
        // current element
        pos = 1 << arr[i];
 
        // Each 0 in _xor represents
        // an even occurrence
        if (!((pos & _xor)!=0))
        {
            // print the even occurring numbers
            System.out.print(arr[i] + " ");
 
            // set bit as 1 to avoid
            // printing duplicates
            _xor ^= pos;
        }
    }
}
 
// Driver code
public static void main(String args[])
{
    int arr[] = {9, 12, 23, 10, 12, 12,
                 15, 23, 14, 12, 15};
    int n = arr.length;
 
    printRepeatingEven(arr, n);
}
}
 
// This code is contributed
// by 29AjayKumar

C#

// C# Program to find the even occurring
// elements in given array
using System;
 
class GFG
{
 
// Function to find the even occurring
// elements in given array
static void printRepeatingEven(int[] arr,
                               int n)
{
    long _xor = 0L;
    long pos;
 
    // do for each element of array
    for (int i = 0; i < n; ++i)
    {
        // left-shift 1 by value of
        // current element
        pos = 1 << arr[i];
 
        // Toggle the bit everytime
        // element gets repeated
        _xor ^= pos;
    }
 
    // Traverse array again and use _xor
    // to find even occurring elements
    for (int i = 0; i < n; ++i)
    {
        // left-shift 1 by value of
        // current element
        pos = 1 << arr[i];
 
        // Each 0 in _xor represents
        // an even occurrence
        if (!((pos & _xor) != 0))
        {
            // print the even occurring numbers
            Console.Write(arr[i] + " ");
 
            // set bit as 1 to avoid
            // printing duplicates
            _xor ^= pos;
        }
    }
}
 
// Driver code
public static void Main()
{
    int[] arr = {9, 12, 23, 10, 12, 12,
                15, 23, 14, 12, 15};
    int n = arr.Length;
 
    printRepeatingEven(arr, n);
}
}
 
// This code is contributed
// by Mukul singh

Python 3

# Python 3 program to find the even
# occurring elements in given array
 
# Function to find the even occurring
# elements in given array
def printRepeatingEven(arr, n):
 
    axor = 0;
 
    # do for each element of array
    for i in range(0, n):
     
        # left-shift 1 by value of
        # current element
        pos = 1 << arr[i];
 
        # Toggle the bit every time
        # element gets repeated
        axor ^= pos;
     
    # Traverse array again and use _xor
    # to find even occurring elements
    for i in range(0, n - 1):
     
        # left-shift 1 by value of
        # current element
        pos = 1 << arr[i];
 
        # Each 0 in _xor represents an
        # even occurrence
        if (not(pos & axor)):
         
            # print the even occurring numbers
            print(arr[i], end = " ");
 
            # set bit as 1 to avoid printing
            # duplicates
            axor ^= pos;
         
# Driver code
arr = [9, 12, 23, 10, 12, 12,
          15, 23, 14, 12, 15 ];
n = len(arr) ;
printRepeatingEven(arr, n);
 
# This code is contributed
# by Shivi_Aggarwal

Javascript

 
// Javascript Program to find the even occurring
// elements in given array
 
// Function to find the even occurring
// elements in given array
function printRepeatingEven(arr, n)
{
    let _xor = 0;
    let pos;
   
    // do for each element of array
    for (let i = 0; i < n; ++i)
    {
        // left-shift 1 by value of
        // current element
        pos = 1 << arr[i];
   
        // Toggle the bit everytime
        // element gets repeated
        _xor ^= pos;
    }
   
    // Traverse array again and use _xor
    // to find even occurring elements
    for (let i = 0; i < n; ++i)
    {
        // left-shift 1 by value of
        // current element
        pos = 1 << arr[i];
   
        // Each 0 in _xor represents
        // an even occurrence
        if (!((pos & _xor)!=0))
        {
            // print the even occurring numbers
            document.write(arr[i] + " ");
   
            // set bit as 1 to avoid
            // printing duplicates
            _xor ^= pos;
        }
    }
}
 
// Driver Code
     
        let arr = [9, 12, 23, 10, 12, 12,
                 15, 23, 14, 12, 15];
        let n = arr.length;
   
        printRepeatingEven(arr, n);
           

输出 ：

12 23 15