检查整数数组的异或是偶数还是奇数

给定一个包含大小为N 的整数的数组arr ，任务是检查这个数组的 XOR 是偶数还是奇数

例子：

Input: arr[] = { 2, 4, 7} 
Output: Odd 
Explanation: 
XOR of array = 2 ^ 4 ^ 7 = 1, which is odd

Input: arr[] = { 3, 9, 12, 13, 15 } 
Output: Even 
 

天真的解决方案：首先找到给定整数数组的异或，然后检查这个异或是偶数还是奇数。
时间复杂度： O(N)

有效的解决方案：更好的解决方案基于位操作事实，即：

• 任意两个偶数或任意两个奇数的按位异或总是偶数。
• 偶数和奇数的按位异或总是奇数。

因此，如果数组中的奇数个数为奇数，则最终的异或将为奇数，如果为偶数，则最终的异或将为偶数。

下面是上述方法的实现：

C++

// C++ program to check if the XOR
// of an array is Even or Odd
 
#include 
using namespace std;
 
// Function to check if the XOR of
// an array of integers is Even or Odd
string check(int arr[], int n)
{
    int count = 0;
 
    for (int i = 0; i < n; i++) {
 
        // Count the number
        // of odd elements
        if (arr[i] & 1)
            count++;
    }
 
    // If count of odd elements
    // is odd, then XOR will be odd
    if (count & 1)
        return "Odd";
 
    // Else even
    else
        return "Even";
}
 
// Driver Code
int main()
{
    int arr[] = { 3, 9, 12, 13, 15 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    // Function call
    cout << check(arr, n) << endl;
 
    return 0;
}

Java

// Java rogram to check if the XOR
// of an array is Even or Odd
import java.util.*;
 
class GFG{
 
// Function to check if the XOR of
// an array of integers is Even or Odd
static String check(int []arr, int n)
{
    int count = 0;
 
    for (int i = 0; i < n; i++) {
 
        // Count the number
        // of odd elements
        if ((arr[i] & 1)!=0)
            count++;
    }
 
    // If count of odd elements
    // is odd, then XOR will be odd
    if ((count & 1)!=0)
        return "Odd";
 
    // Else even
    else
        return "Even";
}
 
// Driver Code
public static void main(String args[])
{
    int []arr = { 3, 9, 12, 13, 15 };
    int n = arr.length;
 
    // Function call
    System.out.println(check(arr, n));
}
}
 
// This code is contributed by Surendra_Gangwar

Python3

# Python3 program to check if the XOR
# of an array is Even or Odd
 
# Function to check if the XOR of
# an array of integers is Even or Odd
def check(arr, n):
    count = 0;
 
    for i in range(n):
 
        # Count the number
        # of odd elements
        if (arr[i] & 1):
            count = count + 1;
     
    # If count of odd elements
    # is odd, then XOR will be odd
    if (count & 1):
        return "Odd";
 
    # Else even
    else:
        return "Even";
 
# Driver Code
if __name__=='__main__':
 
    arr = [ 3, 9, 12, 13, 15 ]
    n = len(arr)
 
    # Function call
    print(check(arr, n))
 
 
# This code is contributed by Princi Singh

C#

// C# program to check if the XOR
// of an array is Even or Odd
using System;
using System.Collections.Generic;
using System.Linq;
  
class GFG
{
 
// Function to check if the XOR of
// an array of integers is Even or Odd
static String check(int []arr, int n)
{
    int count = 0;
 
    for (int i = 0; i < n; i++) {
 
        // Count the number
        // of odd elements
        if (arr[i] == 1)
            count++;
    }
 
    // If count of odd elements
    // is odd, then XOR will be odd
    if (count == 1)
        return "Odd";
 
    // Else even
    else
        return "Even";
}
 
// Driver Code
    public static void Main(String[] args)
    {
    int []arr= { 3, 9, 12, 13, 15 };
    int n = arr.Length;
 
    // Function call
    Console.Write(check(arr, n));
}
}
 
// This code is contributed by shivanisinghss2110

Javascript

 
// Javascript program to check if the XOR
// of an array is Even or Odd
 
// Function to check if the XOR of
// an array of integers is Even or Odd
function check(arr, n)
{
    let count = 0;
 
    for(let i = 0; i < n; i++)
    {
         
        // Count the number
        // of odd elements
        if (arr[i] & 1)
            count++;
    }
 
    // If count of odd elements
    // is odd, then XOR will be odd
    if (count & 1)
        return "Odd";
 
    // Else even
    else
        return "Even";
}
 
// Driver Code
let arr = [ 3, 9, 12, 13, 15 ];
let n = arr.length;
 
// Function call
document.write(check(arr, n));
 
// This code is contributed by subham348
 

输出：

Even

时间复杂度：O(N)

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