子数组的最大长度，以使子数组的总和为偶数

给定N个元素的数组。任务是找到最长子数组的长度，以使子数组的总和为偶数。
例子：

Input : N = 6, arr[] = {1, 2, 3, 2, 1, 4}
Output : 5
Explanation: In the example the subarray 
in range [2, 6] has sum 12 which is even, 
so the length is 5.

Input : N = 4, arr[] = {1, 2, 3, 2}
Output : 4

方法：首先检查数组的总和是否为偶数。如果数组的总和为偶数，则答案为N。
如果数组的总和不是偶数，则表示为奇数。因此，我们的想法是从数组中找到一个奇数元素，以便排除该元素并比较数组两部分的长度，我们可以获得具有偶数和的子数组的最大长度。
很明显，具有偶数和的子数组将存在于[1，x)或(x，N]范围内，
其中1 <= x <= N，而arr [x]是ODD。
下面是上述方法的实现：

C++

// C++ implementation of the above approach
 
#include 
using namespace std;
 
// Function to find length of the longest
// subarray such that sum of the
// subarray is even
int maxLength(int a[], int n)
{
    int sum = 0, len = 0;
 
    // Check if sum of complete array is even
    for (int i = 0; i < n; i++)
        sum += a[i];
 
    if (sum % 2 == 0) // total sum is already even
        return n;
 
    // Find an index i such the a[i] is odd
    // and compare length of both halfs excluding
    // a[i] to find max length subarray
    for (int i = 0; i < n; i++) {
        if (a[i] % 2 == 1)
            len = max(len, max(n - i - 1, i));
    }
 
    return len;
}
 
// Driver Code
int main()
{
    int a[] = { 1, 2, 3, 2 };
    int n = sizeof(a) / sizeof(a[0]);
 
    cout << maxLength(a, n) << "\n";
 
    return 0;
}

Java

// Java implementation of the approach
 
class GFG
{
 
    // Function to find length of the longest
    // subarray such that sum of the
    // subarray is even
    static int maxLength(int a[], int n)
    {
        int sum = 0, len = 0;
 
        // Check if sum of complete array is even
        for (int i = 0; i < n; i++)
        {
            sum += a[i];
        }
 
        if (sum % 2 == 0) // total sum is already even
        {
            return n;
        }
 
        // Find an index i such the a[i] is odd
        // and compare length of both halfs excluding
        // a[i] to find max length subarray
        for (int i = 0; i < n; i++)
        {
            if (a[i] % 2 == 1)
            {
                len = Math.max(len, Math.max(n - i - 1, i));
            }
        }
 
        return len;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int a[] = {1, 2, 3, 2};
        int n = a.length;
        System.out.println(maxLength(a, n));
 
    }
}
 
// This code has been contributed by 29AjayKumar

Python

# Python3 implementation of the above approach
 
# Function to find Length of the longest
# subarray such that Sum of the
# subarray is even
def maxLength(a, n):
 
    Sum = 0
    Len = 0
 
    # Check if Sum of complete array is even
    for i in range(n):
        Sum += a[i]
 
    if (Sum % 2 == 0): # total Sum is already even
        return n
 
    # Find an index i such the a[i] is odd
    # and compare Length of both halfs excluding
    # a[i] to find max Length subarray
    for i in range(n):
        if (a[i] % 2 == 1):
            Len = max(Len, max(n - i - 1, i))
 
    return Len
 
# Driver Code
 
a= [1, 2, 3, 2]
n = len(a)
 
print(maxLength(a, n))
 
# This code is contributed by mohit kumar

C#

// C# implementation of the approach
using System;
 
class GFG
{
     
    // Function to find length of the longest
    // subarray such that sum of the
    // subarray is even
    static int maxLength(int []a, int n)
    {
        int sum = 0, len = 0;
 
        // Check if sum of complete array is even
        for (int i = 0; i < n; i++)
        {
            sum += a[i];
        }
 
        if (sum % 2 == 0) // total sum is already even
        {
            return n;
        }
 
        // Find an index i such the a[i] is odd
        // and compare length of both halfs excluding
        // a[i] to find max length subarray
        for (int i = 0; i < n; i++)
        {
            if (a[i] % 2 == 1)
            {
                len = Math.Max(len, Math.Max(n - i - 1, i));
            }
        }
 
        return len;
    }
 
    // Driver Code
    static public void Main ()
    {
        int []a = {1, 2, 3, 2};
        int n = a.Length;
        Console.WriteLine(maxLength(a, n));
 
    }
}
 
// This code has been contributed by ajit.

Javascript

 
// Javascript implementation of the above approach
 
// Function to find length of the longest
// subarray such that sum of the
// subarray is even
function maxLength(a, n)
{
    let sum = 0, len = 0;
 
    // Check if sum of complete array is even
    for (let i = 0; i < n; i++)
        sum += a[i];
 
    if (sum % 2 == 0) // total sum is already even
        return n;
 
    // Find an index i such the a[i] is odd
    // and compare length of both halfs excluding
    // a[i] to find max length subarray
    for (let i = 0; i < n; i++) {
        if (a[i] % 2 == 1)
            len = Math.max(len, Math.max(n - i - 1, i));
    }
 
    return len;
}
 
// Driver Code
    let a = [ 1, 2, 3, 2 ];
    let n = a.length;
 
    document.write(maxLength(a, n) + "<br>");
 
     
// This code is contributed by Mayank Tyagi
 

输出：

4

时间复杂度： O(N)