数组中两个奇数发生的元素，所有其他偶数发生的时间

给定一个数组，其中所有元素除两次外均出现偶数次，请打印出两个奇数出现的元素。可以假设数组的大小至少为2。

例子：

Input : arr[] = {2, 3, 8, 4, 4, 3, 7, 8}
Output : 2 7

Input : arr[] = {15, 10, 10, 50 7, 5, 5, 50, 50, 50, 50, 50}
Output : 7 15

一个简单的解决方案是使用两个嵌套循环。外循环遍历所有元素。内部循环对当前元素的出现进行计数。我们打印出现次数为奇数的元素。如果此解决方案为O(n ² )，则时间复杂度

更好的解决方案是使用哈希。该解决方案的时间复杂度为O(n)，但需要额外的空间。

一个有效的解决方案是使用按位运算运算符。这个想法是基于在两个缺失要素和两个重复要素中使用的方法。

C++

// CPP code to find two odd occurring elements
// in an array where all other elements appear
// even number of times.
#include 
using namespace std;
  
void printOdds(int arr[], int n)
{
    // Find XOR of all numbers
    int res = 0;
    for (int i = 0; i < n; i++)
        res = res ^ arr[i];
  
    // Find a set bit in the XOR (We find
    // rightmost set bit here)
    int set_bit = res & (~(res - 1));
  
    // Traverse through all numbers and
    // divide them in two groups 
    // (i) Having set bit set at same 
    //     position as the only set bit 
    //     in set_bit
    // (ii) Having 0 bit at same position
    //      as the only set bit in set_bit
    int x = 0, y = 0;
    for (int i = 0; i < n; i++) {
        if (arr[i] & set_bit)
            x = x ^ arr[i];
        else
            y = y ^ arr[i];
    }
  
    // XOR of two different sets are our
    // required numbers.
    cout << x << " " << y;
}
  
// Driver code
int main()
{
    int arr[] = { 2, 3, 3, 4, 4, 5 };
    int n = sizeof(arr) / sizeof(arr[0]);
    printOdds(arr, n);
    return 0;
}

Java

// Java code to find two 
// odd occurring elements
// in an array where all
// other elements appear
// even number of times.
  
class GFG
{
static void printOdds(int arr[], 
                      int n)
{
    // Find XOR of
    // all numbers
    int res = 0;
    for (int i = 0; i < n; i++)
        res = res ^ arr[i];
  
    // Find a set bit in the 
    // XOR (We find rightmost 
    // set bit here)
    int set_bit = res & 
                  (~(res - 1));
  
    // Traverse through all 
    // numbers and divide them 
    // in two groups (i) Having 
    // set bit set at same position 
    // as the only set bit in 
    // set_bit (ii) Having 0 bit at
    // same position as the only 
    // set bit in set_bit
    int x = 0, y = 0;
    for (int i = 0; i < n; i++)
    {
        if ((arr[i] & set_bit) != 0)
            x = x ^ arr[i];
        else
            y = y ^ arr[i];
    }
  
    // XOR of two different 
    // sets are our required
    // numbers.
    System.out.println( x + " " + y);
}
  
// Driver code
public static void main(String [] args)
{
    int arr[] = { 2, 3, 3, 
                  4, 4, 5 };
    int n = arr.length;
    printOdds(arr, n);
}
}
  
// This code is contributed by
// Smitha Dinesh Semwal

Python3

# Python 3 code to find two 
# odd occurring elements in
# an array where all other 
# elements appear even number
# of times.
def printOdds(arr, n):
  
    # Find XOR of all numbers
    res = 0
    for i in range(0, n):
        res = res ^ arr[i]
  
    # Find a set bit in 
    # the XOR (We find 
    # rightmost set bit here)
    set_bit = res & (~(res - 1))
  
    # Traverse through all numbers 
    # and divide them in two groups 
    # (i) Having set bit set at 
    # same position as the only set 
    # bit in set_bit
    # (ii) Having 0 bit at same 
    # position as the only set
    # bit in set_bit
    x = 0
    y = 0
    for i in range(0, n):
        if (arr[i] & set_bit):
            x = x ^ arr[i]
        else:
            y = y ^ arr[i]
      
    # XOR of two different
    # sets are our
    # required numbers.
    print(x , y, end = "")
  
# Driver code
arr = [2, 3, 3, 4, 4, 5 ]
n = len(arr) 
printOdds(arr, n)
  
# This code is contributed
# by Smitha

C#

// C# code to find two 
// odd occurring elements
// in an array where all
// other elements appear
// even number of times.
using System;
  
class GFG
{
static void printOdds(int []arr, 
                      int n)
{
    // Find XOR of
    // all numbers
    int res = 0;
    for (int i = 0; i < n; i++)
        res = res ^ arr[i];
  
    // Find a set bit in the 
    // XOR (We find rightmost 
    // set bit here)
    int set_bit = res & 
               (~(res - 1));
  
    // Traverse through all 
    // numbers and divide them 
    // in two groups (i) Having 
    // set bit set at same position 
    // as the only set bit in 
    // set_bit (ii) Having 0 bit at
    // same position as the only 
    // set bit in set_bit
    int x = 0, y = 0;
    for (int i = 0; i < n; i++)
    {
        if ((arr[i] & set_bit) != 0)
            x = x ^ arr[i];
        else
            y = y ^ arr[i];
    }
  
    // XOR of two different 
    // sets are our required
    // numbers.
    Console.WriteLine(x + " " + y);
}
  
// Driver code
public static void Main()
{
    int []arr = { 2, 3, 3, 
                  4, 4, 5 };
    int n = arr.Length;
    printOdds(arr, n);
}
}
  
// This code is contributed by
// Akanksha Rai(Abby_akku)

输出：

5 2

时间复杂度： O(n)
辅助空间： O(1)