给定一个由N个元素组成的数组arr [] ，任务是找到长度为X的任何子数组的最大和，使得X> 0且X％2 = 0 。
例子：
Input: arr[] = {1, 2, 3}
Output: 5
{2, 3} is the required subarray.
Input: arr[] = {8, 9, -8, 9, 10}
Output: 20
{9, -8, 9, 10} is the required subarray.
Even though {8, 9, -8, 9, 10} has the maximum sum
but it is not of even length.
方法：该问题是最大子数组和问题的一种变体，可以使用动态编程方法解决。创建一个数组dp [] ，其中dp [i]将存储第一个元素为arr [i]的偶数长度子数组的最大和。现在，递归关系将是：
dp[i] = max((arr[i] + arr[i + 1]), (arr[i] + arr[i + 1] + dp[i + 2]))
这是因为以元素arr [i]开头的最大最大偶数长度子数组可以是arr [i]和arr [i + 1]的和，也可以是arr [i] + arr [i + 1]的和以arr [i + 2]即dp [i + 2]开头的偶数长度子数组的最大和。取两者中的最大值。
最后，来自dp []数组的最大值将是必需的答案。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
// Function to return the maximum
// subarray sum of even length
int maxEvenLenSum(int arr[], int n)
{
// There has to be at
// least 2 elements
if (n < 2)
return 0;
// dp[i] will store the maximum
// subarray sum of even length
// starting at arr[i]
int dp[n] = { 0 };
// Valid subarray cannot start from
// the last element as its
// length has to be even
dp[n - 1] = 0;
dp[n - 2] = arr[n - 2] + arr[n - 1];
for (int i = n - 3; i >= 0; i--) {
// arr[i] and arr[i + 1] can be added
// to get an even length subarray
// starting at arr[i]
dp[i] = arr[i] + arr[i + 1];
// If the sum of the valid subarray starting
// from arr[i + 2] is greater than 0 then it
// can be added with arr[i] and arr[i + 1]
// to maximize the sum of the subarray
// starting from arr[i]
if (dp[i + 2] > 0)
dp[i] += dp[i + 2];
}
// Get the sum of the even length
// subarray with maximum sum
int maxSum = *max_element(dp, dp + n);
return maxSum;
}
// Driver code
int main()
{
int arr[] = { 8, 9, -8, 9, 10 };
int n = sizeof(arr) / sizeof(int);
cout << maxEvenLenSum(arr, n);
return 0;
} Java
// Java implementation of the approach
import java.util.Arrays;
class GFG
{
// Function to return the maximum
// subarray sum of even length
static int maxEvenLenSum(int arr[], int n)
{
// There has to be at
// least 2 elements
if (n < 2)
return 0;
// dp[i] will store the maximum
// subarray sum of even length
// starting at arr[i]
int []dp = new int[n];
// Valid subarray cannot start from
// the last element as its
// length has to be even
dp[n - 1] = 0;
dp[n - 2] = arr[n - 2] + arr[n - 1];
for (int i = n - 3; i >= 0; i--)
{
// arr[i] and arr[i + 1] can be added
// to get an even length subarray
// starting at arr[i]
dp[i] = arr[i] + arr[i + 1];
// If the sum of the valid subarray starting
// from arr[i + 2] is greater than 0 then it
// can be added with arr[i] and arr[i + 1]
// to maximize the sum of the subarray
// starting from arr[i]
if (dp[i + 2] > 0)
dp[i] += dp[i + 2];
}
// Get the sum of the even length
// subarray with maximum sum
int maxSum = Arrays.stream(dp).max().getAsInt();
return maxSum;
}
// Driver code
public static void main(String[] args)
{
int arr[] = { 8, 9, -8, 9, 10 };
int n = arr.length;
System.out.println(maxEvenLenSum(arr, n));
}
}
// This code is contributed by 29AjayKumarPython3
# Python3 implementation of the approach
# Function to return the maximum
# subarray sum of even length
def maxEvenLenSum(arr, n):
# There has to be at
# least 2 elements
if (n < 2):
return 0
# dp[i] will store the maximum
# subarray sum of even length
# starting at arr[i]
dp = [0 for i in range(n)]
# Valid subarray cannot start from
# the last element as its
# length has to be even
dp[n - 1] = 0
dp[n - 2] = arr[n - 2] + arr[n - 1]
for i in range(n - 3, -1, -1):
# arr[i] and arr[i + 1] can be added
# to get an even length subarray
# starting at arr[i]
dp[i] = arr[i] + arr[i + 1]
# If the sum of the valid subarray
# starting from arr[i + 2] is
# greater than 0 then it can be added
# with arr[i] and arr[i + 1]
# to maximize the sum of the
# subarray starting from arr[i]
if (dp[i + 2] > 0):
dp[i] += dp[i + 2]
# Get the sum of the even length
# subarray with maximum sum
maxSum = max(dp)
return maxSum
# Driver code
arr = [8, 9, -8, 9, 10]
n = len(arr)
print(maxEvenLenSum(arr, n))
# This code is contributed by Mohit KumarC#
// C# implementation of the approach
using System;
class GFG
{
static int MaxSum(int []arr)
{
// assigning first element to the array
int large = arr[0];
// loop to compare value of large
// with other elements
for (int i = 1; i < arr.Length; i++)
{
// if large is smaller than other element
// assig that element to the large
if (large < arr[i])
large = arr[i];
}
return large;
}
// Function to return the maximum
// subarray sum of even length
static int maxEvenLenSum(int []arr, int n)
{
// There has to be at
// least 2 elements
if (n < 2)
return 0;
// dp[i] will store the maximum
// subarray sum of even length
// starting at arr[i]
int []dp = new int[n];
// Valid subarray cannot start from
// the last element as its
// length has to be even
dp[n - 1] = 0;
dp[n - 2] = arr[n - 2] + arr[n - 1];
for (int i = n - 3; i >= 0; i--)
{
// arr[i] and arr[i + 1] can be added
// to get an even length subarray
// starting at arr[i]
dp[i] = arr[i] + arr[i + 1];
// If the sum of the valid subarray starting
// from arr[i + 2] is greater than 0 then it
// can be added with arr[i] and arr[i + 1]
// to maximize the sum of the subarray
// starting from arr[i]
if (dp[i + 2] > 0)
dp[i] += dp[i + 2];
}
// Get the sum of the even length
// subarray with maximum sum
int maxSum = MaxSum(dp);
return maxSum;
}
// Driver code
public static void Main()
{
int []arr = { 8, 9, -8, 9, 10 };
int n = arr.Length;
Console.WriteLine(maxEvenLenSum(arr, n));
}
}
// This code is contributed by kanugargng输出：
20
时间复杂度： O(n)
空间复杂度： O(n)