最大和交替子阵列

给定一个大小为N的数组arr[] ，任务是找到给定数组可能的子数组的最大交替和。

 Alternating Subarray Sum: Considering a subarray {arr[i], arr[j]}, alternating sum of the subarray is arr[i] – arr[i + 1] + arr[i + 2] – …….. (+ / -) arr[j].

例子：

Input: arr[] = {-4, -10, 3, 5}
Output: 9
Explanation: Subarray {arr[0], arr[2]} = {-4, -10, 3}. Therefore, the sum of this subarray is 9.

Input: arr[] = {-1, 2, -1, 4, 7}
Output: 7

方法：给定的问题可以通过使用动态规划来解决。请按照以下步骤解决问题：

• 初始化一个变量，比如sum为0 ，它将保存一个最大的交替子数组 sum 和一个变量，比如sumSoFar ，用于存储从^第一个循环中的偶数索引开始的子数组的总和和从奇数索引开始的总和，在第二个循环中^nd循环。
• 在两个循环的每次迭代中，将sum更新为max(sum, sumSoFar) 。
• 最后，返回存储在sum变量中的最大交替和。

下面是上述方法的实现：

C++

// C++ implementation for the above approach
#include 
using namespace std;
 
// Function to find the maximum alternating
// sum of a subarray for the given array
int alternatingSum(int arr[],int n)
{
  int sum = 0;
  int sumSoFar = 0;
 
  // Traverse the array
  for (int i = 0; i < n; i++) {
 
    // Store sum of subarrays
    // starting at even indices
    if (i % 2 == 1) {
 
      sumSoFar -= arr[i];
    }
    else {
 
      sumSoFar = max(
        sumSoFar + arr[i], arr[i]);
    }
 
    // Update sum
    sum = max(sum, sumSoFar);
  }
 
  sumSoFar = 0;
 
  // Traverse the array
  for (int i = 1; i < n; i++) {
 
    // Store sum of subarrays
    // starting at odd indices
    if (i % 2 == 0) {
      sumSoFar -= arr[i];
    }
    else {
      sumSoFar = max(
        sumSoFar + arr[i], arr[i]);
    }
 
    // Update sum
    sum = max(sum, sumSoFar);
  }
  return sum;
}
 
// Driver code
int main()
{
 
  // Given Input
  int arr[] ={ -4, -10, 3, 5 };
  int n = sizeof(arr)/sizeof(arr[0]);
   
  // Function call
  int ans = alternatingSum(arr,n);
 
  cout<

---

Java

// Java implementation for the above approach
 
import java.io.*;
 
class GFG {
 
    // Function to find the maximum alternating
    // sum of a subarray for the given array
    public static int alternatingSum(int[] arr)
    {
        int sum = 0;
        int sumSoFar = 0;
 
        // Traverse the array
        for (int i = 0; i < arr.length; i++) {
 
            // Store sum of subarrays
            // starting at even indices
            if (i % 2 == 1) {
 
                sumSoFar -= arr[i];
            }
            else {
 
                sumSoFar = Math.max(
                    sumSoFar + arr[i], arr[i]);
            }
 
            // Update sum
            sum = Math.max(sum, sumSoFar);
        }
 
        sumSoFar = 0;
 
        // Traverse the array
        for (int i = 1; i < arr.length; i++) {
 
            // Store sum of subarrays
            // starting at odd indices
            if (i % 2 == 0) {
                sumSoFar -= arr[i];
            }
            else {
                sumSoFar = Math.max(
                    sumSoFar + arr[i], arr[i]);
            }
 
            // Update sum
            sum = Math.max(sum, sumSoFar);
        }
        return sum;
    }
 
    // Driver code
    public static void main(String[] args)
    {
 
        // Given Input
        int arr[] = new int[] { -4, -10, 3, 5 };
 
        // Function call
        int ans = alternatingSum(arr);
 
        System.out.println(ans);
    }
}

Python3

# Python implementation for the above approach
 
# Function to find the maximum alternating
# sum of a subarray for the given array
def alternatingSum(arr, n):
    sum_ = 0
    sumSoFar = 0
     
    # Traverse the array
    for i in range(n):
       
      # Store sum of subarrays
       # starting at even indices
        if i % 2 == 1:
            sumSoFar -= arr[i]
        else:
            sumSoFar = max(arr[i], sumSoFar + arr[i])
             
            # Update sum
        sum_ = max(sum_, sumSoFar)
 
    sumSoFar = 0
     
    # Traverse array
    for i in range(1, n):
       
      # Store sum of subarrays
      # starting at odd indices
        if i % 2 == 0:
            sumSoFar -= arr[i]
        else:
            sumSoFar = max(sumSoFar, sumSoFar + arr[i])
        sum_ = max(sum_, sumSoFar)
         
        # update sum
    return sum_
 
# given array
arr = [-4, -10, 3, 5]
n = len(arr)
 
# return sum
ans = alternatingSum(arr, n)
print(ans)
 
# This code is contributed by Parth Manchanda

C#

// C# implementation for the above approach
using System;
using System.Collections.Generic;
 
class GFG{
 
// Function to find the maximum alternating
// sum of a subarray for the given array
static int alternatingSum(int []arr,int n)
{
    int sum = 0;
    int sumSoFar = 0;
     
    // Traverse the array
    for(int i = 0; i < n; i++)
    {
     
        // Store sum of subarrays
        // starting at even indices
        if (i % 2 == 1)
        {
            sumSoFar -= arr[i];
        }
        else
        {
            sumSoFar = Math.Max(
            sumSoFar + arr[i], arr[i]);
        }
         
        // Update sum
        sum = Math.Max(sum, sumSoFar);
    }
     
    sumSoFar = 0;
     
    // Traverse the array
    for(int i = 1; i < n; i++)
    {
     
        // Store sum of subarrays
        // starting at odd indices
        if (i % 2 == 0)
        {
            sumSoFar -= arr[i];
        }
        else
        {
            sumSoFar = Math.Max(
            sumSoFar + arr[i], arr[i]);
        }
         
        // Update sum
        sum = Math.Max(sum, sumSoFar);
    }
    return sum;
}
 
// Driver code
public static void Main()
{
     
    // Given Input
    int []arr = { -4, -10, 3, 5 };
    int n = arr.Length;
     
    // Function call
    int ans = alternatingSum(arr,n);
     
    Console.Write(ans);
}
}
 
// This code is contributed by SURENDRA_GANGWAR

Javascript

// JavaScript implementation for the above approach
    // Function to find the maximum alternating
    // sum of a subarray for the given array
    function alternatingSum(arr)
    {
        var sum = 0;
        var sumSoFar = 0;
 
        // Traverse the array
        for (var i = 0; i < arr.length; i++) {
 
            // Store sum of subarrays
            // starting at even indices
            if (i % 2 == 1) {
 
                sumSoFar -= arr[i];
            }
            else {
 
                sumSoFar = Math.max(
                    sumSoFar + arr[i], arr[i]);
            }
 
            // Update sum
            sum = Math.max(sum, sumSoFar);
        }
 
        sumSoFar = 0;
 
        // Traverse the array
        for (var i = 1; i < arr.length; i++) {
 
            // Store sum of subarrays
            // starting at odd indices
            if (i % 2 == 0) {
                sumSoFar -= arr[i];
            }
            else {
                sumSoFar = Math.max(
                    sumSoFar + arr[i], arr[i]);
            }
 
            // Update sum
            sum = Math.max(sum, sumSoFar);
        }
        return sum;
    }
 
    // Driver code
        // Given Input
        var arr = new Array ( -4, -10, 3, 5 );
 
        // Function call
        var ans = alternatingSum(arr);
        document.write(ans);
         
    // This code is contributed by  shivanisinghss2110

输出：

9

时间复杂度： O(N)
辅助空间： O(1)

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