找出两条对角线之和之间的差异

给定一个n X n的矩阵。任务是计算其对角线之和之间的绝对差。
例子：

Input : mat[][] = 11 2 4
                   4 5 6
                  10 8 -12 
Output : 15
Sum of primary diagonal = 11 + 5 + (-12) = 4.
Sum of secondary diagonal = 4 + 5 + 10 = 19.
Difference = |19 - 4| = 15.


Input : mat[][] = 10 2
                   4 5
Output : 7

计算一个方阵的两条对角线的和。沿着矩阵的第一条对角线，行索引 = 列索引，即如果 i = j，则 mat[i][j] 位于第一条对角线上。沿着另一条对角线，如果 i = n-1-j，行索引 = n – 1 – 列索引，即 mat[i][j] 位于第二条对角线上。通过使用两个循环，我们遍历整个矩阵并计算矩阵对角线的总和。
下面是这种方法的实现：

C++

// C++ program to find the difference
// between the sum of diagonal.
#include 
#define MAX 100
using namespace std;
 
int difference(int arr[][MAX], int n)
{
    // Initialize sums of diagonals
    int d1 = 0, d2 = 0;
 
    for (int i = 0; i < n; i++)
    {
        for (int j = 0; j < n; j++)
        {
            // finding sum of primary diagonal
            if (i == j)
                d1 += arr[i][j];
 
            // finding sum of secondary diagonal
            if (i == n - j - 1)
                d2 += arr[i][j];
        }
    }
 
    // Absolute difference of the sums
    // across the diagonals
    return abs(d1 - d2);
}
 
// Driven Program
int main()
{
    int n = 3;
 
    int arr[][MAX] =
    {
        {11, 2, 4},
        {4 , 5, 6},
        {10, 8, -12}
    };
 
    cout << difference(arr, n);
    return 0;
}

Java

// JAVA Code for Find difference between sums
// of two diagonals
class GFG {
     
    public static int difference(int arr[][], int n)
    {
        // Initialize sums of diagonals
        int d1 = 0, d2 = 0;
      
        for (int i = 0; i < n; i++)
        {
            for (int j = 0; j < n; j++)
            {
                // finding sum of primary diagonal
                if (i == j)
                    d1 += arr[i][j];
      
                // finding sum of secondary diagonal
                if (i == n - j - 1)
                    d2 += arr[i][j];
            }
        }
      
        // Absolute difference of the sums
        // across the diagonals
        return Math.abs(d1 - d2);
    }
     
    /* Driver program to test above function */
    public static void main(String[] args)
    {
        int n = 3;
          
        int arr[][] =
        {
            {11, 2, 4},
            {4 , 5, 6},
            {10, 8, -12}
        };
      
        System.out.print(difference(arr, n));
        
    }
  }
// This code is contributed by Arnav Kr. Mandal.

Python3

# Python3 program to find the difference
# between the sum of diagonal.
def difference(arr, n):
 
    # Initialize sums of diagonals
    d1 = 0
    d2 = 0
 
    for i in range(0, n):
     
        for j in range(0, n):
         
            # finding sum of primary diagonal
            if (i == j):
                d1 += arr[i][j]
 
            # finding sum of secondary diagonal
            if (i == n - j - 1):
                d2 += arr[i][j]
         
    # Absolute difference of the sums
    # across the diagonals
    return abs(d1 - d2);
 
# Driver Code
n = 3
 
arr = [[11, 2, 4],
       [4 , 5, 6],
       [10, 8, -12]]
 
print(difference(arr, n))
     
# This code is contributed
# by ihritik

C#

// C# Code for find difference between
// sums of two diagonals
using System;
 
public class GFG
{
 
    // Function to calculate difference
    public static int difference(int[,] arr,
                                 int n)
    {
         
        // Initialize sums of diagonals
        int d1 = 0, d2 = 0;
     
        for (int i = 0; i < n; i++)
        {
            for (int j = 0; j < n; j++)
            {
                 
                // finding sum of primary diagonal
                if (i == j)
                    d1 += arr[i, j];
     
                // finding sum of secondary diagonal
                if (i == n - j - 1)
                    d2 += arr[i, j];
            }
        }
     
        // Absolute difference of the
        // sums across the diagonals
        return Math.Abs(d1 - d2);
    }
     
    // Driver Code
    public static void Main()
    {
        int n = 3;
         
        int[,] arr ={{11, 2, 4},
                     {4 , 5, 6},
                     {10, 8, -12}};
     
        Console.Write(difference(arr, n));
         
    }
}
 
// This code is contributed by shiv_bhakt.

Javascript

 
// Javascript Code for Find difference between sums
// of two diagonals
 
function difference(arr,n)
    {
        // Initialize sums of diagonals
        let d1 = 0, d2 = 0;
     
        for (let i = 0; i < n; i++)
        {
            for (let j = 0; j < n; j++)
            {
                // finding sum of primary diagonal
                if (i == j)
                    d1 += arr[i][j];
     
                // finding sum of secondary diagonal
                if (i == n - j - 1)
                    d2 += arr[i][j];
            }
        }
     
        // Absolute difference of the sums
        // across the diagonals
        return Math.abs(d1 - d2);
    }
     
    /* Driver program to test above function */
     
        let n = 3;
         
        let arr =
        [
            [11, 2, 4],
            [4 , 5, 6],
            [10, 8, -12]
        ];
     
        document.write(difference(arr, n));
     
// This code is contributed Bobby
 

输出：

15

时间复杂度：O(n*n)
我们可以使用单元格索引中存在的模式优化上述解决方案以在 O(n) 中工作。

C++

// C++ program to find the difference
// between the sum of diagonal.
#include 
#define MAX 100
using namespace std;
 
int difference(int arr[][MAX], int n)
{
    // Initialize sums of diagonals
    int d1 = 0, d2 = 0;
 
    for (int i = 0; i < n; i++)
    {
        d1 += arr[i][i];
        d2 += arr[i][n-i-1];
    }
 
    // Absolute difference of the sums
    // across the diagonals
    return abs(d1 - d2);
}
 
// Driven Program
int main()
{
    int n = 3;
 
    int arr[][MAX] =
    {
        {11, 2, 4},
        {4 , 5, 6},
        {10, 8, -12}
    };
 
    cout << difference(arr, n);
    return 0;
}

Java

// JAVA Code for Find difference between sums
// of two diagonals
 
class GFG {
     
    public static int difference(int arr[][], int n)
    {
        // Initialize sums of diagonals
        int d1 = 0, d2 = 0;
      
        for (int i = 0; i < n; i++)
        {
            d1 += arr[i][i];
            d2 += arr[i][n-i-1];
        }
      
        // Absolute difference of the sums
        // across the diagonals
        return Math.abs(d1 - d2);
    }
     
    /* Driver program to test above function */
    public static void main(String[] args)
    {
        int n = 3;
          
        int arr[][] =
        {
            {11, 2, 4},
            {4 , 5, 6},
            {10, 8, -12}
        };
      
        System.out.print(difference(arr, n));
        
    }
  }
// This code is contributed by Arnav Kr. Mandal.

Python3

# Python3 program to find the difference
# between the sum of diagonal.
def difference(arr, n):
 
    # Initialize sums of diagonals
    d1 = 0
    d2 = 0
 
    for i in range(0, n):
        d1 = d1 + arr[i][i]
        d2 = d2 + arr[i][n - i - 1]
         
    # Absolute difference of the sums
    # across the diagonals
    return abs(d1 - d2)
 
# Driver Code
n = 3
 
arr = [[11, 2, 4],
       [4 , 5, 6],
       [10, 8, -12]]
 
print(difference(arr, n))
     
# This code is contributed
# by ihritik

C＃

// C# Code for find difference between
// sums of two diagonals
using System;
 
public class GFG
 
{
     
    //Function to find difference
    public static int difference(int[,] arr,
                                 int n)
    {
         
        // Initialize sums of diagonals
        int d1 = 0, d2 = 0;
     
        for (int i = 0; i < n; i++)
        {
            d1 += arr[i, i];
            d2 += arr[i, n - i - 1];
        }
     
        // Absolute difference of the sums
        // across the diagonals
        return Math.Abs(d1 - d2);
    }
     
    // Driver Code
    public static void Main()
    {
        int n = 3;
         
        int[,] arr ={{11, 2, 4},
                     {4 , 5, 6},
                     {10, 8, -12}};
     
        Console.Write(difference(arr, n));
         
    }
}
 
// This code is contributed by shiv_bhakt.

Javascript

// JAVA SCRIPT  Code for Find difference between sums
// of two diagonals
 
function difference(arr,n)
    {
     
        // Initialize sums of diagonals
        let d1 = 0, d2 = 0;
     
        for (let i = 0; i < n; i++)
        {
            d1 += arr[i][i];
            d2 += arr[i][n-i-1];
        }
     
        // Absolute difference of the sums
        // across the diagonals
        return Math.abs(d1 - d2);
    }
     
    /* Driver program to test above function */
     
        let n = 3;       
        let arr = [[11, 2, 4],
            [4 , 5, 6],
            [10, 8, -12]];
     
        document.write(difference(arr, n));
         
 
// This code is contributed by sravan kumar Gottumukkala

输出：

15