在给定数组中找到最大的连续对和

给定大小为N的整数数组arr [] ，任务是找到连续对{a，b} ，以使该对中两个元素的总和最大。如果不止一对这样的对具有最大和，则打印其中任何一对。如果有多对具有最大和，请打印其中的任何一对。
例子：

Input: arr[] = {1, 2, 3, 4} 
Output: 3 4 
Explanation: 
Here, the contiguous pairs in the array are: 
{1, 2} -> Sum 3 
{2, 3} -> Sum 5 
{3, 4} -> Sum 7 
Maxiumum sum is 7 for the pair is (3, 4) so this is answer.

Input: arr[] = {11, -5, 9, -3, 2} 
Output: 11 -5 
The contiguous pairs with their respective sums are : 
{11, -5} -> Sum 6 
{-5, 9} -> Sum 4 
{9, -3} -> Sum 6 
{-3, 2} -> Sum -1 
The maximum sum obtained is 6 from the pairs (11, -5) and (9, -3). 
 

方法：
请按照以下步骤解决问题：

1. 逐一生成所有连续对并计算总和。
2. 将每个对的和与最大和进行比较，并相应地更新与最大和相对应的对。
3. 返回代表最大和的对。

下面是上述方法的实现：

C++

// C++ program to find the 
// a contiguous pair from the
// which has the largest sum
#include 
using namespace std;
  
// Function to find and return
// the largest sum contiguous pair
vector largestSumpair(int arr[], int n)
{
      
    // Stores the contiguous pair 
    vector pair;
      
    // Intialize maximum sum
    int max_sum = INT_MIN, i;
      
    for(i = 1; i < n; i++) 
    {
          
        // Compare sum of pair with max_sum
        if (max_sum < (arr[i] + arr[i - 1]))
        {
            max_sum = arr[i] + arr[i - 1];
            if (pair.empty()) 
            { 
                  
                // Insert the pair
                pair.push_back(arr[i - 1]);
                pair.push_back(arr[i]);
            }
            else
            {
                pair[0] = arr[i - 1];
                pair[1] = arr[i];
            }
        }
        return pair;
    }
}
      
// Driver Code 
int main() 
{ 
    int arr[] = {11, -5, 9, -3, 2};
    int N = sizeof(arr) / sizeof(arr[0]);
      
    vector pair = largestSumpair(arr, N);
    for(auto it = pair.begin(); it != pair.end(); ++it)
    {
        cout << *it << ' '; 
    }
      
    return 0; 
} 
  
// This code is contributed by shubhamsingh10

Java

// Java program to find the 
// a contiguous pair from the 
// which has the largest sum 
import java.util.*;
  
class GFG{
      
// Function to find and return 
// the largest sum contiguous pair 
public static Vector largestSumpair(int[] arr,
                                             int n) 
{ 
      
    // Stores the contiguous pair 
    Vector pair = new Vector(); 
      
    // Intialize maximum sum 
    int max_sum = Integer.MIN_VALUE, i; 
      
    for(i = 1; i < n; i++) 
    { 
          
        // Compare sum of pair with max_sum 
        if (max_sum < (arr[i] + arr[i - 1])) 
        { 
            max_sum = arr[i] + arr[i - 1]; 
              
            if (pair.isEmpty()) 
            { 
                  
                // Insert the pair 
                pair.add(arr[i - 1]); 
                pair.add(arr[i]); 
            } 
            else
            { 
                pair.set(0, arr[i - 1]);
                pair.set(1, arr[i]);
            } 
        } 
    } 
    return pair;
} 
  
// Driver Code    
public static void main(String[] args) 
{
    int arr[] = { 11, -5, 9, -3, 2 }; 
    int N = arr.length; 
      
    Vector pair = new Vector(); 
    pair = largestSumpair(arr, N);
    for (Integer it : pair)
    {         
        System.out.print(it + " ");
    }
}
}
  
// This code is contributed by divyeshrabadiya07

Python3

# Python3 program to find the 
# a contiguous pair from the
# which has the largest sum
  
# importing sys
import sys
  
# Function to find and return
# the largest sum contiguous pair
def largestSumpair(arr, n):
# Stores the contiguous pair 
    pair = []
  
# Intialize maximum sum
    max_sum = -sys.maxsize-1
  
    for i in range(1, n):
      
        # Compare sum of pair with max_sum
        if max_sum < ( arr[i] + arr[i-1] ):
            max_sum = arr[i] + arr[i-1]
          
            if pair == []:
                # Insert the pair
                pair.append(arr[i-1])
                pair.append(arr[i])
            else:
                pair[0] = arr[i-1]
                pair[1] = arr[i]
  
    return pair
      
      
# Driver Code 
arr = [11, -5, 9, -3, 2] 
N = len(arr) 
pair = largestSumpair(arr, N)
print(pair[0], end =" ")
print(pair[1], end =" ")

C#

// C# program to find the 
// a contiguous pair from the 
// which has the largest sum 
using System;
using System.Collections; 
using System.Collections.Generic;
  
class GFG{
  
// Function to find and return 
// the largest sum contiguous pair 
public static ArrayList largestSumpair(int[] arr,
                                       int n) 
{ 
      
    // Stores the contiguous pair 
    ArrayList pair = new ArrayList(); 
      
    // Intialize maximum sum 
    int max_sum = int.MinValue, i; 
      
    for(i = 1; i < n; i++) 
    { 
          
        // Compare sum of pair with max_sum 
        if (max_sum < (arr[i] + arr[i - 1])) 
        { 
            max_sum = arr[i] + arr[i - 1]; 
              
            if (pair.Count == 0) 
            { 
                  
                // Insert the pair 
                pair.Add(arr[i - 1]); 
                pair.Add(arr[i]); 
            } 
            else
            { 
                pair[0] = arr[i - 1];
                pair[1] = arr[i];
            } 
        } 
    } 
    return pair;
} 
  
// Driver code
public static void Main(string[] args)
{
    int []arr = { 11, -5, 9, -3, 2 }; 
    int N = arr.Length; 
      
    ArrayList pair = new ArrayList(); 
    pair = largestSumpair(arr, N);
      
    foreach(int it in pair)
    {         
        Console.Write(it + " ");
    }
}
}
  
// This code is contributed by rutvik_56

输出：

11 -5

时间复杂度： O(N)
辅助空间： O(1)