高效计算矩阵对角线和的Java程序

给定一个二维方阵，求主对角线和次对角线中元素的总和。例如，考虑以下 4 X 4 输入矩阵。

A00 A01 A02 A03
A10 A11 A12 A13
A20 A21 A22 A23
A30 A31 A32 A33

主对角线由元素 A00、A11、A22、A33 形成。

1. 主对角线条件：行列条件是行 = 列。
   次对角线由元素 A03、A12、A21、A30 形成。
2. 次对角线的条件：行列条件是 row = numberOfRows – column -1。

例子 ：

Input : 
4
1 2 3 4
4 3 2 1
7 8 9 6
6 5 4 3
Output :
Principal Diagonal: 16
Secondary Diagonal: 20

Input :
3
1 1 1
1 1 1
1 1 1
Output :
Principal Diagonal: 3
Secondary Diagonal: 3

方法 1 (O(n ^ 2) ：

在这种方法中，我们使用两个循环，即一个循环用于列和一个循环用于行，并且在内部循环中，我们检查上述条件：

Java

// A simple java program to find
// sum of diagonals
import java.io.*;
  
public class GFG {
  
    static void printDiagonalSums(int [][]mat,
                                         int n)
    {
        int principal = 0, secondary = 0;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
      
                // Condition for principal
                // diagonal
                if (i == j)
                    principal += mat[i][j];
      
                // Condition for secondary
                // diagonal
                if ((i + j) == (n - 1))
                    secondary += mat[i][j];
            }
        }
      
        System.out.println("Principal Diagonal:"
                                    + principal);
                                      
        System.out.println("Secondary Diagonal:"
                                    + secondary);
    }
  
    // Driver code
    static public void main (String[] args)
    {
          
        int [][]a = { { 1, 2, 3, 4 },
                      { 5, 6, 7, 8 }, 
                      { 1, 2, 3, 4 },
                      { 5, 6, 7, 8 } };
                      
        printDiagonalSums(a, 4);
    }
}
  
// This code is contributed by vt_m.

输出：

Principal Diagonal:18
Secondary Diagonal:18

此代码需要 O(n^2) 时间和 O(1) 辅助空间

方法2（O（n）：

在这种方法中，我们使用一个循环，即一个用于计算主对角线和次对角线之和的循环：

Java

// An efficient java program to find
// sum of diagonals
import java.io.*;
  
public class GFG {
  
    static void printDiagonalSums(int [][]mat,
                                        int n)
    {
        int principal = 0, secondary = 0; 
        for (int i = 0; i < n; i++) {
            principal += mat[i][i];
            secondary += mat[i][n - i - 1]; 
        }
      
        System.out.println("Principal Diagonal:"
                                   + principal);
                                     
        System.out.println("Secondary Diagonal:"
                                   + secondary);
    }
      
    // Driver code
    static public void main (String[] args)
    {
        int [][]a = { { 1, 2, 3, 4 },
                      { 5, 6, 7, 8 }, 
                      { 1, 2, 3, 4 },
                      { 5, 6, 7, 8 } };
      
        printDiagonalSums(a, 4);
    }
}
  
// This code is contributed by vt_m.

输出 ：

Principal Diagonal:18
Secondary Diagonal:18

此代码需要 O(n) 时间和 O(1) 辅助空间
有关详细信息，请参阅有关有效计算矩阵对角线和的完整文章！