给定数组arr [] ，任务是通过反转子数组并打印获得的最大和来最大化偶数索引元素的和。
例子：
Input: arr[] = {1, 2, 1, 2, 1}
Output: 5
Explanation:
Sum of intial even-indexed elements = a[0] + a[2] + a[4] = 1 + 1 + 1 = 3
Reversing subarray {1, 2, 1, 2} modifies the array to {2, 1, 2, 1, 1}.
Hence, the maximized sum = 2 + 2 + 1 = 5
Input: arr[] = {7, 8, 4, 5, 7, 6, 8, 9, 7, 3}
Output: 37
天真的方法：
解决问题的最简单方法是通过元素的一一反转生成所有可能的排列，并为每个排列计算偶数索引的总和。打印所有排列中的最大可能和。
时间复杂度： O(N 3 )
辅助空间： O(N)
高效方法：
通过使用动态编程来检查阵列旋转的最大差异，可以将上述方法进一步优化为O(N)计算复杂度。
请按照以下步骤解决问题：
- 比较奇数索引和偶数索引的元素，并跟踪它们。
- 初始化两个数组leftDP []和rightDP [] 。
- 对于每个奇数索引， leftDP []存储元素在当前索引处与左边元素的差， rightDP []存储元素在当前索引处的差。
- 如果为前一个索引计算的差为正，则将其添加到当前差中：
if(dp[i – 1] > 0)
dp[i] = dp[i-1] + curr_diff
- 否则，存储当前差异：
dp[i] = curr_diff;
下面是上述方法的实现：
C++
// C++ program to implement
// the above approach
#include
using namespace std;
// Function to return maximized sum
// at even indices
int maximizeSum(int arr[], int n)
{
int sum = 0;
for(int i = 0; i < n; i = i + 2)
sum += arr[i];
// Stores difference with
// element on the left
int leftDP[n / 2];
// Stores difference with
// element on the right
int rightDP[n / 2];
int c = 0;
for(int i = 1; i < n; i = i + 2)
{
// Compute and store
// left difference
int leftDiff = arr[i] - arr[i - 1];
// For first index
if (c - 1 < 0)
leftDP = leftDiff;
else
{
// If previous difference
// is positive
if (leftDP > 0)
leftDP = leftDiff + leftDP;
// Otherwise
else
leftDP[i] = leftDiff;
}
int rightDiff;
// For the last index
if (i + 1 >= n)
rightDiff = 0;
// Otherwise
else
rightDiff = arr[i] - arr[i + 1];
// For first index
if (c - 1 < 0)
rightDP = rightDiff;
else
{
// If the previous difference
// is positive
if (rightDP > 0)
rightDP = rightDiff +
rightDP;
else
rightDP = rightDiff;
}
c++;
}
int maxi = 0;
for(int i = 0; i < n / 2; i++)
{
maxi = max(maxi, max(leftDP[i],
rightDP[i]));
}
return maxi + sum;
}
// Driver Code
int main()
{
int arr[] = { 7, 8, 4, 5, 7,
6, 8, 9, 7, 3 };
int n = sizeof(arr) / sizeof(arr[0]);
int ans = maximizeSum(arr, n);
cout << (ans);
}
// This code is contributed by chitranayal Java
// Java Program to implement
// the above approach
import java.io.*;
class GFG {
// Function to return maximized sum
// at even indices
public static int maximizeSum(int[] arr)
{
int n = arr.length;
int sum = 0;
for (int i = 0; i < n; i = i + 2)
sum += arr[i];
// Stores difference with
// element on the left
int leftDP[] = new int[n / 2];
// Stores difference with
// element on the right
int rightDP[] = new int[n / 2];
int c = 0;
for (int i = 1; i < n; i = i + 2) {
// Compute and store
// left difference
int leftDiff = arr[i]
- arr[i - 1];
// For first index
if (c - 1 < 0)
leftDP = leftDiff;
else {
// If previous difference
// is positive
if (leftDP > 0)
leftDP = leftDiff
+ leftDP;
// Otherwise
else
leftDP[i] = leftDiff;
}
int rightDiff;
// For the last index
if (i + 1 >= arr.length)
rightDiff = 0;
// Otherwise
else
rightDiff = arr[i]
- arr[i + 1];
// For first index
if (c - 1 < 0)
rightDP = rightDiff;
else {
// If the previous difference
// is positive
if (rightDP > 0)
rightDP = rightDiff
+ rightDP;
else
rightDP = rightDiff;
}
c++;
}
int max = 0;
for (int i = 0; i < n / 2; i++) {
max = Math.max(max,
Math.max(
leftDP[i],
rightDP[i]));
}
return max + sum;
}
// Driver Code
public static void main(String[] args)
{
int arr[] = { 7, 8, 4, 5, 7, 6,
8, 9, 7, 3 };
int ans = maximizeSum(arr);
System.out.println(ans);
}
}Python3
# Python3 program to implement
# the above approach
# Function to return maximized sum
# at even indices
def maximizeSum(arr):
n = len(arr)
sum = 0
for i in range(0, n, 2):
sum += arr[i]
# Stores difference with
# element on the left
leftDP = [0] * (n)
# Stores difference with
# element on the right
rightDP = [0] * (n)
c = 0
for i in range(1, n, 2):
# Compute and store
# left difference
leftDiff = arr[i] - arr[i - 1]
# For first index
if (c - 1 < 0):
leftDP[i] = leftDiff
else:
# If previous difference
# is positive
if (leftDP[i] > 0):
leftDP[i] = (leftDiff +
leftDP[i - 1])
# Otherwise
else:
leftDP[i] = leftDiff
rightDiff = 0
# For the last index
if (i + 1 >= len(arr)):
rightDiff = 0
# Otherwise
else:
rightDiff = arr[i] - arr[i + 1]
# For first index
if (c - 1 < 0):
rightDP[i] = rightDiff
else:
# If the previous difference
# is positive
if (rightDP[i] > 0):
rightDP[i] = (rightDiff +
rightDP[i - 1])
else:
rightDP[i] = rightDiff
c += 1
maxm = 0
for i in range(n // 2):
maxm = max(maxm, max(leftDP[i],
rightDP[i]))
return maxm + sum
# Driver Code
if __name__ == '__main__':
arr = [ 7, 8, 4, 5, 7,
6, 8, 9, 7, 3 ]
ans = maximizeSum(arr)
print(ans)
# This code is contributed by mohit kumar 29C#
// C# program to implement
// the above approach
using System;
class GFG{
// Function to return maximized sum
// at even indices
public static int maximizeSum(int[] arr)
{
int n = arr.Length;
int sum = 0;
for(int i = 0; i < n; i = i + 2)
sum += arr[i];
// Stores difference with
// element on the left
int []leftDP = new int[n / 2];
// Stores difference with
// element on the right
int []rightDP = new int[n / 2];
int c = 0;
for(int i = 1; i < n; i = i + 2)
{
// Compute and store
// left difference
int leftDiff = arr[i] - arr[i - 1];
// For first index
if (c - 1 < 0)
leftDP = leftDiff;
else
{
// If previous difference
// is positive
if (leftDP > 0)
leftDP = leftDiff +
leftDP;
// Otherwise
else
leftDP = leftDiff;
}
int rightDiff;
// For the last index
if (i + 1 >= arr.Length)
rightDiff = 0;
// Otherwise
else
rightDiff = arr[i] - arr[i + 1];
// For first index
if (c - 1 < 0)
rightDP = rightDiff;
else
{
// If the previous difference
// is positive
if (rightDP > 0)
rightDP = rightDiff +
rightDP;
else
rightDP = rightDiff;
}
c++;
}
int max = 0;
for(int i = 0; i < n / 2; i++)
{
max = Math.Max(max,
Math.Max(leftDP[i],
rightDP[i]));
}
return max + sum;
}
// Driver Code
public static void Main(String[] args)
{
int []arr = { 7, 8, 4, 5, 7, 6,
8, 9, 7, 3 };
int ans = maximizeSum(arr);
Console.WriteLine(ans);
}
}
// This code is contributed by 29AjayKumar输出：
37
时间复杂度： O(N)
辅助空间： O(N)