给定一个大小为N的整数数组arr[] ，任务是找到连续对{a, b}使得该对中两个元素的总和最大。如果有不止一个这样的对的最大总和，则打印任何这样的对。在多对的总和最大的情况下，打印其中任何一对。
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：
- 一一生成所有连续对并计算总和。
- 将每对的总和与最大总和进行比较，并相应地更新与最大总和相对应的对。
- 返回代表最大总和的对。
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)
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