Java程序查找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 primary 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：

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.

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。

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.

输出：

15

时间复杂度： O(n)
有关详细信息，请参阅有关查找两条对角线之和之间的差异的完整文章！