使用负数最大化集合的两个子集之间的差异

给定一个大小为 N 的整数。任务是将这些整数分成两组 g1 和 g2，使得 (g1 的元素总和) - (g2 的元素总和) 变为最大值。你的任务是打印结果的值。我们可以保留一个子集为空。
例子：

Input : 3, 7, -4, 10, -11, 2 
Output : 37 
Explanation: 
g1: 3, 7, 10, 2 
g2: -4, -11 
result = ( 3 + 7 + 10 + 2 ) – ( -4 + -11) = 22 – (-15) = 37
Input : 2, 2, -2, -2 
Output : 8 

这个想法是根据它们的符号值对整数进行分组，即我们将正整数分组为g1，将负整数分组为g2。
因为， – ( -g2 ) = +g2
因此，结果变为 g1 + |g2|。

C++

// CPP program to make two subsets with
// maximum difference.
#include 
using namespace std;
 
int maxDiff(int arr[], int n)
{
    int sum = 0;
 
    // We move all negative elements into
    // one set. So we add negation of negative
    // numbers to maximize difference
    for (int i = 0; i < n; i++)
         sum = sum + abs(arr[i]);
    
    return sum;
}
 
// Driver Code
int main()
{
    int arr[] = { 3, 7, -4, 10, -11, 2 };
    int n = sizeof(arr)/sizeof(arr[0]);
    cout << maxDiff(arr, n);
    return 0;
}

Java

// Java program to make two subsets with
// maximum difference.
import java.util.*;
 
class solution
{
 
static int maxDiff(int arr[], int n)
{
    int sum = 0;
 
    // We move all negative elements into
    // one set. So we add negation of negative
    // numbers to maximize difference
    for (int i = 0; i < n; i++)
        sum = sum + Math.abs(arr[i]);
     
    return sum;
}
 
// Driver Code
public static void main(String args[])
{
    int []arr = { 3, 7, -4, 10, -11, 2 };
    int n = arr.length;
    System.out.println(maxDiff(arr, n));
}
}

Python3

# Python3 program to make two subsets
# with maximum difference.
 
def maxDiff(arr, n) :
 
    sum = 0
 
    # We move all negative elements into
    # one set. So we add negation of negative
    # numbers to maximize difference
    for i in range(n) :
        sum += abs(arr[i])
     
    return sum
 
# Driver Code
if __name__ == "__main__" :
 
    arr = [ 3, 7, -4, 10, -11, 2 ]
    n = len(arr)
    print(maxDiff(arr, n))
 
# This code is contributed by Ryuga

C#

using System;
 
// C# program to make two subsets with
// maximum difference.
 
public class solution
{
 
public static int maxDiff(int[] arr, int n)
{
    int sum = 0;
 
    // We move all negative elements into
    // one set. So we add negation of negative
    // numbers to maximize difference 
    for (int i = 0; i < n; i++)
    {
        sum = sum + Math.Abs(arr[i]);
    }
 
    return sum;
}
 
// Driver Code
public static void Main(string[] args)
{
    int[] arr = new int[] {3, 7, -4, 10, -11, 2};
    int n = arr.Length;
    Console.WriteLine(maxDiff(arr, n));
}
}
 
  // This code is contributed by Shrikant13

Javascript

// javascript program to make two subsets with
// maximum difference.
function maxDiff(arr , n)
{
    var sum = 0;
 
    // We move all negative elements into
    // one set. So we add negation of negative
    // numbers to maximize difference
    for (i = 0; i < n; i++)
        sum = sum + Math.abs(arr[i]);
     
    return sum;
}
 
// Driver Code
var arr = [ 3, 7, -4, 10, -11, 2 ];
var n = arr.length;
document.write(maxDiff(arr, n));
 
// This code is contributed by Princi Singh

输出：

37

时间复杂度： O(n)