求严格递增子数组的最大和

给定一个正整数数组。求严格递增子数组的最大和。请注意，此问题不同于最大子数组和和最大和增加子序列问题。

例子：

Input : arr[] = {1, 2, 3, 2, 5, 1, 7}
Output : 8
Explanation : Some Strictly increasing subarrays are
{1, 2, 3} sum = 6,
{2, 5} sum = 7,
{1, 7} sum 8
Maximum Sum = 8

Input : arr[] = {1, 2, 2, 4}
Output: 6
Explanation : Increasing subarray with maximum sum is 6.

编程需要懂一点英语

一个简单的解决方案是生成所有可能的子数组，并为每个子数组检查子数组是否严格增加。如果子数组是严格递增的，那么我们计算 sum 并更新 max_sum。时间复杂度 O(n ² )。

这个问题的有效解决方案需要 O(n) 时间。这个想法是跟踪最大总和和当前总和。对于每个元素 arr[i]，如果它大于 arr[i-1]，那么我们将它添加到当前总和中。否则 arr[i] 是最大和增加子数组的另一个潜在候选者的起点，因此我们将当前和更新为数组。但是在更新当前总和之前，如果需要，我们会更新最大总和。

Let input array be 'arr[]' and size of array be 'n'

Initialize : 
max_sum = arr[0]// because if array size is 1 than it would return that element.
 // used to store the maximum sum 
current_sum = arr[0] // used to compute current sum 

// Traverse array starting from second element
i goes from 1 to n-1

    // Check if it is strictly increasing then we 
    // update current_sum.
    current_sum = current_sum + arr[i]
    max_sum = max(max_sum, current_sum)// Also needed for subarray having last element.
    // else strictly increasing subarray breaks and 
    // arr[i] is starting point of next potential
    // subarray
    max_sum = max(max_sum, current_sum)
    current_sum = arr[i]

return max(max_sum, current_sum)

下面是上述思想的实现。

C++

// C/C++ program to find the maximum sum of strictly
// increasing subarrays
#include 
using namespace std;
 
// Returns maximum sum of strictly increasing
// subarrays
int maxsum_SIS(int arr[], int n)
{
    // Initialize max_sum be 0
    int max_sum = arr[0];
 
    // Initialize current sum be arr[0]
    int current_sum = arr[0];
 
    // Traverse array elements after first element.
    for (int i = 1; i < n; i++)
    {
        // update current_sum for
        // strictly increasing subarray
        if (arr[i - 1] < arr[i])
        {
            current_sum = current_sum + arr[i];
            max_sum = max(max_sum, current_sum);
        }
 
        else // strictly increasing subarray break
        {
            // update max_sum and current_sum ;
            max_sum = max(max_sum, current_sum);
            current_sum = arr[i];
        }
    }
 
    return max(max_sum, current_sum);
}
 
// Driver code
int main()
{
    int arr[] = { 1, 2, 2, 4 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    cout << "Maximum sum : " << maxsum_SIS(arr, n);
    return 0;
}

Java

// Java program to find the
// maximum sum of strictly increasing subarrays
 
public class GFG {
 
    // Returns maximum sum
    // of strictly increasing subarrays
    static int maxsum_SIS(int arr[], int n)
    {
        // Initialize max_sum be 0
        int max_sum = arr[0];
 
        // Initialize current sum be arr[0]
        int current_sum = arr[0];
 
        // Traverse array elements after first element.
        for (int i = 1; i < n; i++)
        {
            // update current_sum
            // for strictly increasing subarray
            if (arr[i - 1] < arr[i])
            {
                current_sum = current_sum + arr[i];
                max_sum = Math.max(max_sum, current_sum);
            }
            else // strictly increasing subarray break
            {
                // update max_sum and current_sum ;
                max_sum = Math.max(max_sum, current_sum);
                current_sum = arr[i];
            }
        }
 
        return Math.max(max_sum, current_sum);
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int arr[] = { 1, 2, 2, 4 };
        int n = arr.length;
        System.out.println("Maximum sum : "
                           + maxsum_SIS(arr, n));
    }
}
 
// This code is contributed by 29AjayKumar

Python3

# Python3 program to find the maximum sum of strictly
# increasing subarrays
 
# Returns maximum sum of strictly increasing
# subarrays
 
 
def maxsum_SIS(arr, n):
    # Initialize max_sum be 0
    max_sum = arr[0]
 
    # Initialize current sum be arr[0]
    current_sum = arr[0]
 
    # Traverse array elements after first element.
    for i in range(1, n):
        # update current_sum for strictly increasing subarray
        if (arr[i-1] < arr[i]):
            current_sum = current_sum + arr[i]
            max_sum = max(max_sum, current_sum)
 
        else:
            # strictly increasing subarray break
            # update max_sum and current_sum
            max_sum = max(max_sum, current_sum)
            current_sum = arr[i]
 
    return max(max_sum, current_sum)
 
# Driver code
 
def main():
    arr = [1, 2, 2, 4]
    n = len(arr)
 
    print("Maximum sum : ", maxsum_SIS(arr, n)),
 
 
if __name__ == '__main__':
    main()
 
# This code is contributed by 29AjayKumar

C#

// C# program to find the maximum sum of strictly
// increasing subarrays
using System;
public class GFG {
 
    // Returns maximum sum of strictly increasing
    // subarrays
    static int maxsum_SIS(int[] arr, int n)
    {
        // Initialize max_sum be 0
        int max_sum = arr[0];
 
        // Initialize current sum be arr[0]
        int current_sum = arr[0];
 
        // Traverse array elements after first element.
        for (int i = 1; i < n; i++) {
            // update current_sum for strictly increasing
            // subarray
            if (arr[i - 1] < arr[i]) {
                current_sum = current_sum + arr[i];
                max_sum = Math.Max(max_sum, current_sum);
            }
            else // strictly increasing subarray break
            {
                // update max_sum and current_sum ;
                max_sum = Math.Max(max_sum, current_sum);
                current_sum = arr[i];
            }
        }
 
        return Math.Max(max_sum, current_sum);
    }
 
    // Driver code
    public static void Main()
    {
        int[] arr = { 1, 2, 2, 4 };
        int n = arr.Length;
        Console.WriteLine("Maximum sum : "
                          + maxsum_SIS(arr, n));
    }
}
 
// This code is contributed by 29AjayKumar

PHP

Javascript

输出

Maximum sum : 6

时间复杂度： O(n)
辅助空间： O(1)